1、IP地址统计问题
注： 有些小细节会导致比较难排查的错误，例如ip字符串和mask字符串分离是否真确，即使一个字母，也会出错。
#include
#include
#include
int toNum(char *str_p, unsigned char *out)//需要保证out为大于或等于4个元素的数组
{
int i;
int pos;
int cnt;
char tmp[128] = {0};
char *tmp_p = tmp;
i = 0;
pos = 0;
cnt = 0;
while(str_p[i] != '\0')
{
if(str_p[i] == '.')
{
tmp_p = str_p + pos;
tmp_p[i - pos] = '\0';
out[cnt] = atoi(tmp_p);
if(out[cnt] == 0 && tmp_p[0] != '0')
return -1;
pos = i + 1;
cnt++;
if(cnt >= 3)
break;
}
i++;
}
if(cnt != 3)
return -1;
tmp_p = str_p + pos;
out[3] = atoi(tmp_p);
return 0;
}
int maskIsOk(unsigned char *num_mask)
{
int i, j;
int mark;
int one_num;
mark = 0;
one_num = 0; //全部为1或者0的掩码为非法
for(i = 0; i < 4; i++)
{
for(j = 0;j < 8; j++)
{
if((num_mask[i] & (1 << (7 - j))) != 0)
{
one_num++;
if(mark == 1)
{
return -1;
}
}
else if(mark == 0)
{
mark = 1;
continue;
}
}
}
if(one_num == 0 || one_num >= 32)
{
return -1;
}
return 0;
}
int isIpPrivate(unsigned char *num_ip)
{
int ip;
int type;
int i;
ip = num_ip[0] << 24 | num_ip[1] << 16 | num_ip[2] << 8 | num_ip[3];
const int start[] =
{
10 << 24,
172 << 24 | 16 << 16,
192 << 24 | 168 << 16
};
const int end[] =
{
10 << 24 | 255 << 16 | 255 << 8 |255,
172 << 24 | 31 << 16 | 255 << 8 | 255 ,
192 << 24 | 168 << 16 | 255 << 8 | 255
};
for(i = 0; i < 3; i++)
{
if(ip >= start[i] && ip <= end[i])
{
return 0;
}
}
return -1;
}
int ipType(unsigned char *num_ip)
{
int ip;
int type;
int i;
ip = num_ip[0] << 24 | num_ip[1] << 16 | num_ip[2] << 8 | num_ip[3];
if(num_ip[0] == 0 || num_ip[0] == 127)
{
return 0;
}
const int start[] =
{
1 << 24,
128 << 24,
192 << 24,
224 << 24,
240 << 24
};
const int end[] =
{
126 << 24 | 255 << 16 | 255 << 8 |255,
191 << 24 | 255 << 16 | 255 << 8 |255,
223 << 24 | 255 << 16 | 255 << 8 |255,
239 << 24 | 255 << 16 | 255 << 8 |255,
255 << 24 | 255 << 16 | 255 << 8 |255
};
for(i = 0; i < 5; i++)
{
if(ip > start[i] && ip < end[i])
{
return (i + 1);
}
}
return -1;
}
int main(void)
{
int i, j;
char str_in[128] = {0};
char str_ip[128] = {0};
char str_mask[128] = {0};
char *str_ip_p;
char *str_mask_p;
unsigned char num_ip[4] = {0};
unsigned char num_mask[4] = {0};
int a, b, c, d, e, error, pri;
a = b = c = d = e = error = pri = 0;
int ret, type;
while(scanf("%s", str_in) != EOF)
{
strcpy(str_ip, str_in);
strcpy(str_mask, str_in);
i = 0;
while(str_in[i] != '\0')
{
if(str_in[i] == '~')
{
str_ip_p = str_ip;
str_ip_p[i] = '\0';
str_mask_p = str_mask + i + 1;
break;
}
i++;
}
ret = toNum(str_mask_p, num_mask);
if(ret != 0)
{
error++;
continue;
}
else if(maskIsOk(num_mask) != 0)
{
error++;
continue;
}
ret = toNum(str_ip_p, num_ip);
if(ret != 0)
{
error++;
continue;
}
else
{
type = ipType(num_ip);
if(type == 1)
a++;
else if(type == 2)
b++;
else if(type == 3)
c++;
else if(type == 4)
d++;
else if(type == 5)
e++;
else if(type == 0)
;
else
{error++; continue;}
if(isIpPrivate(num_ip) == 0)
pri++;
}
}
printf("%d %d %d %d %d %d %d\n", a, b, c, d, e, error, pri);
return 0;
}
2、密码验证合格程序
#include
#include
int main(void)
{
char psswd[128] = {0};
int len;
int mark[4];
int brk;
int i, j;
while(scanf("%s", psswd) != EOF)
{
//printf("----------------------->\n");
len = strlen(psswd);
if(len <= 8)
{
printf("%s\n", "NG");
continue;
}
mark[0] = mark[1] = mark[2] = mark[3] = 0;
for(i = 0; i < len; i ++)
{
if((psswd[i] >= 'a' && psswd[i] <= 'z'))
{
mark[0] = 1;
}
else if((psswd[i] >= 'A' && psswd[i] <= 'Z'))
{
mark[1] = 1;
}
else if((psswd[i] >= '0' && psswd[i] <= '9'))
{
mark[2] = 1;
}
else
{
mark[3] = 1;
}
}
if((mark[0] + mark[1] + mark[2] + mark[3]) < 3)
{
printf("%s\n", "NG");
continue;
}
brk = 0;
for(i = 0; i < len - 5; i ++)
{
if(brk == 1)
{
printf("%s\n", "NG");
break;
}
for(j = i + 3; j < len - 2; j++)
{
if(psswd[i] == psswd[j] && psswd[i + 1] == psswd[j + 1] && psswd[i + 2] == psswd[j + 2])
{
brk = 1;
break;
}
}
}
if(brk == 1)
{
continue;
}
printf("%s\n", "OK");
}
return 0;
}
3、求四则式子的数值
前提： 输入的格式是正确的，不考虑对格式的判断
例子： 1+2*10-(-2+3)
/*注：（1）括号里的式子用递归解决；
*（2）存储+-符号用sybol[0]，存储* /符号用symbol[1]；存储第一个计数值用num[0]，第二个数值用num[1]，主要是为了解决乘除的优先级高于加减这个规则；
*（3）遇到符号时，对前面的数值计算；
*/
#include
#include
#define PRINT_EN 0
int mypow(int n, int cnt)
{
int i;
int ret = 1;
for(i = 0; i < cnt; i++)
{
ret *= n;
}
return ret;
}
int calc(char *str_p, int set_pos)
{
int symbol[2] = {0};
int num[2] = {0};
int tmp[16] = {0};
int tmp_value, result;
char ch;
tmp_value = result = 0;
static int pos = 0; //位置
if(set_pos == 1)
{
pos = 0;
}
int mark;
mark = 0;
while(1)
{
if(mark == 1)
ch = '+';
else if(str_p[pos] == '\0')//字符串末尾多个 +，执行最后操作
ch = '+';
else
ch = str_p[pos];
if(ch >= '0' && ch <= '9')
{
tmp_value *= 10;
tmp_value += ch - '0';
}
else if(ch == '+' || ch == '-')
{
if(symbol[1] == 1)//上一个为乘法
{
num[1] *= tmp_value;
if(symbol[0] == 1 || symbol[0] == 0)//上上个为加法
{
num[0] += num[1];
}
else if(symbol[0] == 2)//上上个为减法法
{
num[0] -= num[1];
}
}
else if(symbol[1] == 2)//上一个为除法
{
if(tmp_value != 0)
{
num[1] /= tmp_value;
}
if(symbol[0] == 1)//上上个为加法
{
num[0] += num[1];
}
else if(symbol[0] == 2)//上上个为减法法
{
num[0] -= num[1];
}
}
else if(symbol[1] == 0 && (symbol[0] == 1 || symbol[0] == 0))
{
num[0] += tmp_value;
}
else if(symbol[1] == 0 && symbol[0] == 2)
{
num[0] -= tmp_value;
}
symbol[1] = 0;
if(ch == '+')
{
if(PRINT_EN)printf("(+)===>num0=%d num1=%d\n", num[0], num[1]);
symbol[0] = 1;
}
else if(ch == '-')
{
if(PRINT_EN)printf("(-)===>num0=%d num1=%d\n", num[0], num[1]);
symbol[0] = 2;
}
tmp_value = 0;
}
else if(ch == '*' || ch == '/')
{
if(symbol[1] == 1)
{
num[1] *= tmp_value;
}
else if(symbol[1] == 2)
{
num[1] /= tmp_value;
}
else if(symbol[1] == 0 )
{
num[1] = tmp_value;
}
if(ch == '*')
{
if(PRINT_EN)printf("(*)===>num0=%d num1=%d\n", num[0], num[1]);
symbol[1] = 1;
}
else if(ch == '/')
{
if(PRINT_EN)printf("(/)===>num0=%d num1=%d\n", num[0], num[1]);
symbol[1] = 2;
}
tmp_value = 0;
}
else if(ch == '(' || ch == '[' || ch == '{')
{
pos++;
if(PRINT_EN)printf("(digui)------------------\n");
tmp_value = calc(str_p, 0);
if(PRINT_EN)printf("(digui)+++++++++++ str_p:%c pos:%d tmp_value:%d\n", *str_p, pos, tmp_value);
}
else if(ch == ')' || ch == ']' || ch == '}')
{
mark = 1;
continue;
}
if(mark == 1)
break;
if(str_p[pos] == '\0')
break;
else
pos++;
}
return num[0];
}
int main(void)
{
//char str[] = "-3-(5+6)*(3-1)";
//char str[] = "3+1";
char str[512] = {0};
scanf("%s", str);
printf("%d\n", calc(str, 1));
return 0;
}
参考
#include
#include
int pos;
int compute(char* data)
{
int len = strlen(data);
int stack[1000];
int top = -1;
char flag = '+';
int num = 0;
while(pos < len)
{
if(data[pos] == '{' || data[pos] == '[' || data[pos] == '(')
{
pos++;
num = compute(data);
}
while(pos < len && isdigit(data[pos]))
{
num = 10 * num + data[pos] - '0';
pos++;
}
switch(flag)
{
case '+':
stack[++top] = num;
break;
case '-':
stack[++top] = -num;
break;
case '*':
stack[top] *= num;
break;
case '/':
stack[top] /= num;
break;
}
num = 0;
flag = data[pos];
if(data[pos] == '}' || data[pos] == ']' || data[pos] == ')')
{
pos++;
break;
}
pos++;
}
int res = 0;
for(int i = 0; i <= top; i++)
{
res += stack[i];
}
return res;
}
int main()
{
char data[1000];
while(scanf("%s", data) != EOF)
{
pos = 0;
int res = compute(data);
printf("%d\n",res);
}
return 0;
}
4、图片整理
注：大小排序
//冒泡法
#include
#include
int main(void)
{
char buf[1024] = {0};
int len;
int i, j;
while(scanf("%s", buf) != EOF)
{
len = strlen(buf);
for(i = 0; i < len - 1; i++)
{
for(j = 0; j < (len - i - 1); j++)
{
if(buf[j] > buf[j + 1])
{
buf[j] ^= buf[j + 1];
buf[j + 1] ^= buf[j];
buf[j] ^= buf[j + 1];
}
}
}
printf("%s\n", buf);
}
return 0;
}
快排法移植
#include
#include
void swap(char *x, char *y) {
char t = *x;
*x = *y;
*y = t;
}
void quick_sort_recursive(char arr[], int start, int end) {
if (start >= end)
return;
int mid = arr[end];
int left = start, right = end - 1;
while (left < right) {
while (arr[left] < mid && left < right)
left++;
while (arr[right] >= mid && left < right)
right--;
swap(&arr[left], &arr[right]);
}
if (arr[left] >= arr[end])
swap(&arr[left], &arr[end]);
else
left++;
if (left)
quick_sort_recursive(arr, start, left - 1);
quick_sort_recursive(arr, left + 1, end);
}
void quick_sort(char arr[], int len) {
quick_sort_recursive(arr, 0, len - 1);
}
int main(void)
{
char buf[1024] = {0};
int len;
int i, j;
while(scanf("%s", buf) != EOF)
{
len = strlen(buf);
quick_sort(buf, len);
printf("%s\n", buf);
}
return 0;
}
参考网友的：位图标记法
#include
#include
int main(){
char s[1025];
while (scanf("%s", s)!= EOF){
int a[1025] ={0};
int len=strlen(s);
for(int i=0; i< len; i++)
{
a[s[i]]++;
}
for(int i=0; i<1025; i++)
{
while(a[i]!=0){
printf("%c",i);
a[i]--;
}
}
printf("\n");
}
}
5、蛇形矩阵
#include
int getNum(int num_start, int step_start, int idx)
{
int i;
int tmp, step;
tmp = num_start;
step = step_start;
for(i = 0; i < idx; i++)
{
tmp += step;
step++;
}
return tmp;
}
int main(void)
{
int i, j;
int num;
while(scanf("%d", &num) != EOF)
{
for(i = 0; i < num; i++)
{
for(j = 0; j < num - i; j++)
{
printf("%d ", getNum(getNum(1, 1, i), 2 + i, j));
}
printf("\n");
}
}
return 0;
}
或者
#include
int main(void)
{
int i, j;
int num;
int step = 1;
int start = 1;
int start2 = 2;
int step3 = 2;
int start3 = 1;
while(scanf("%d", &num) != EOF)
{
for(i = 0; i < num; i++)
{
for(j = 0; j < num - i; j++)
{
if(j == 0)
{
printf("%d ", start3);
}
else
{
printf("%d ", start3 + step3);
start3 += step3;
step3++;
}
}
printf("\n");
start += step;
step++;
start3 = start;
start2++;
step3 = start2;
}
}
return 0;
}
6、砝码称重
6.1 未通过（原因超时；空间规模小的话，还可以计算）
分析： 用的是穷举法，二进制为1代表数存在，0代表不存在，2的n次方遍历代表n个数的组合个数。我发现，最近刷的题用这种方法来解决总是耗时久，效率不好。
//注： 遍历次数超过2 的 30 次方，耗时愈加明显，达到秒的级别
#include
#include
#define MAX_GROUP_NUM 10
#define MAX_FAMA_NUM 6
#define MAXFAMA_WIGHT 2000
int strToNumArray(char *str, int *array, int *array_n, int max_num)
{
int i, j, tmp, cnt;
char buf[16] = {0};
i = j = cnt = 0;
while(str[i] != '\0')
{
if(str[i] == ' ')
{
i++;
continue;
}
tmp = i;
while(str[i] != ' ' && str[i] != '\0')
{
buf[i - tmp] = str[i];
i++;
}
buf[i - tmp] = '\0';
array[cnt] = atoi(buf);
if(array[cnt] == 0 && buf[0] != '0')
{
*array_n = cnt;
return -1;
}
cnt++;
if(cnt >= max_num)
{
*array_n = cnt;
return -1;
}
}
*array_n = cnt;
return 0;
}
int main(void)
{
long long long_i, long_j;
int i, j;
int val[MAX_GROUP_NUM] = {0};
int num[MAX_GROUP_NUM] = {0};
int val_all[MAX_FAMA_NUM * MAX_GROUP_NUM] = {0};
int group, totle;
char str_val[64] = {0};
char str_num[64] = {0};
int val_n, num_n;
int bit_map[MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1] = {0};
int tmp, cnt;
while(scanf("%d\n", &group) != EOF)
{
gets(str_val);
gets(str_num);
strToNumArray(str_val, val, &val_n, MAX_GROUP_NUM);
strToNumArray(str_num, num, &num_n, MAX_GROUP_NUM);
totle = 0;
for(i = 0; i < group; i ++)
{
for(j = 0; j < num[i]; j++)
{
val_all[totle] = val[i];
totle++;
}
}
/* for(i = 0; i < group; i ++)
{
printf("%d ", val[i]);
}
printf("\n");
for(i = 0; i < group; i ++)
{
printf("%d ", num[i]);
}
printf("\n");
for(i = 0; i < totle; i ++)
{
printf("%d ", val_all[i]);
}
printf("\n");*/
for(i = 0; i < MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1; i++)
bit_map[i] = 0;
for(long_i = 0; long_i < (((long long)1) << totle); long_i++)
{
tmp = 0;
//printf("===>");
for(j = 0; j < totle; j++)
{
if((long_i & ((long long)1 << j)) != 0)
{
//printf("[j=%d]vla:%d ", j, val_all[j]);
tmp += val_all[j];
}
}
//printf("<==i:0x%x tmp:%d\n", long_i, tmp);
bit_map[tmp] = 1;
}
cnt = 0;
for(i = 0; i < MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1; i++)
{
if(bit_map[i] == 1)
{
cnt++;
}
}
printf("%d\n", cnt);
}
return 0;
}
6.2 通过：递归法，耗时110ms
注：遇到错误的点：数组溢出
#include
#include
#define MAX_GROUP_NUM 10
#define MAX_FAMA_NUM 6
#define MAXFAMA_WIGHT 2000
int strToNumArray(char *str, int *array, int *array_n, int max_num)
{
int i, j, tmp, cnt;
char buf[16] = {0};
i = j = cnt = 0;
while(str[i] != '\0')
{
if(str[i] == ' ')
{
i++;
continue;
}
tmp = i;
while(str[i] != ' ' && str[i] != '\0')
{
buf[i - tmp] = str[i];
i++;
}
buf[i - tmp] = '\0';
array[cnt] = atoi(buf);
if(array[cnt] == 0 && buf[0] != '0')
{
*array_n = cnt;
return -1;
}
cnt++;
if(cnt >= max_num)
{
*array_n = cnt;
return -1;
}
}
*array_n = cnt;
return 0;
}
//多重循环
int record[100];
int bit_map[MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1] ;
int forLoop(int *end, int n, int top, int *data)
{
int i;
int tmp;
if(end == NULL || data == NULL || top < n)return -1;
if(n == 0)
{
tmp = 0;
for(i = 0; i < top - n; i++)
{
tmp += data[i] * record[i];
}
bit_map[tmp] = 1;
}
else
{
for(i = 0; i <= end[top - n]; i++)
{
record[top - n] = i;
forLoop(end, n - 1, top, data);
}
}
return 0;
}
int main(void)
{
int val[MAX_GROUP_NUM] = {0};
int num[MAX_GROUP_NUM] = {0};
char str_val[64] = {0};
char str_num[64] = {0};
int val_n, num_n, cnt, i, group;
while(scanf("%d\n", &group) != EOF)
{
gets(str_val);
gets(str_num);
strToNumArray(str_val, val, &val_n, MAX_GROUP_NUM);
strToNumArray(str_num, num, &num_n, MAX_GROUP_NUM);
for(i = 0; i < MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1; i++)
bit_map[i] = 0;
forLoop(num, group, group, val);
cnt = 0;
for(i = 0; i < MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1; i++)
{
if(bit_map[i] == 1)
{
cnt++;
}
}
printf("%d\n", cnt);
}
return 0;
}
6.3 通过，类似于动态规划：利用前者的结果推导后面的结果，耗时10ms
#include
#include
#include
#define MAX_GROUP_NUM 10
#define MAX_FAMA_NUM 6
#define MAXFAMA_WIGHT 2000
int strToNumArray(char *str, int *array, int *array_n, int max_num)
{
int i, j, tmp, cnt;
char buf[16] = {0};
i = j = cnt = 0;
while(str[i] != '\0')
{
if(str[i] == ' ')
{
i++;
continue;
}
tmp = i;
while(str[i] != ' ' && str[i] != '\0')
{
buf[i - tmp] = str[i];
i++;
}
buf[i - tmp] = '\0';
array[cnt] = atoi(buf);
if(array[cnt] == 0 && buf[0] != '0')
{
*array_n = cnt;
return -1;
}
cnt++;
if(cnt >= max_num)
{
*array_n = cnt;
return -1;
}
}
*array_n = cnt;
return 0;
}
int main(void)
{
int val[MAX_GROUP_NUM + 2] = {0};
int num[MAX_GROUP_NUM + 2] = {0};
char str_val[64] = {0};
char str_num[64] = {0};
int val_n, num_n, cnt, i, j, k, group;
int bit_map[MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1] ;
int bit_map_tmp[MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1] ;
while(scanf("%d\n", &group) != EOF)
{
gets(str_val);
gets(str_num);
strToNumArray(str_val, val, &val_n, MAX_GROUP_NUM);
strToNumArray(str_num, num, &num_n, MAX_GROUP_NUM);
for(i = 0; i < MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1; i++)
bit_map[i] = 0;
cnt = 0;
for(i = 0; i < group; i++)
{
memcpy(bit_map_tmp, bit_map, sizeof(bit_map));
for(j = 0; j < MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1; j++)
{
if(bit_map_tmp[j] == 1)
{
for(k = 0; k <= num[i]; k++)
{
bit_map[val[i] * k + j] = 1;
}
}
}
for(j = 0; j <= num[i]; j++)
{
bit_map[val[i] * j] = 1;
}
}
for(i = 0; i < MAXFAMA_WIGHT * MAX_FAMA_NUM * MAX_GROUP_NUM + 1; i++)
{
if(bit_map[i] == 1)
{
cnt++;
}
}
printf("%d\n", cnt);
}
return 0;
}
6.4 参考网友：只要3ms
我的分析：
（1）位图 1 代表存在， 0代表不存在
（2）以上一个数的结果为基础，推导下一个数的结果，例如
求{2， 2， 3， 3， 3} 和的不同情况：
1）0放到位图，2放到位图===>[0,2]
2）2放到位图，0+2放到位图；2+2放到位图===>[0,2,4]
3） 3、0+3 、2+3、4+3 放到位图 ===>[0,2,4,3,5,7]
以此类推，类似于数学的 “演绎推理”。
#include
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int i,j,k,m[n],x[n],max=0,count=0;
for(i=0;i=0;k--)
if(array[k]==1)
array[k+m[i]]=1;
for(i=0;i<=max;i++)
if(array[i]==1)
count++;
printf("%d\n",count);
}
return 0;
}
7、学英文（英文数字）
#include
#include
const char name_of_num[][32]={
"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten",
"eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen","eighteen", "nineteen"
};
const char name_of_numplus[][32]={
"aa", "aa", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"
};
const char name_key[][32]={
"and", "hundred", "thousand", "million", "billon"
};
void getNamePre(int num, char (*out_str)[32], int *pos)
{
int tmp;
if(num >= 1000) return ;
tmp = num / 100;
if(tmp >= 1)
{
strcpy(out_str[(*pos)++], name_of_num[tmp]);
strcpy(out_str[(*pos)++], name_key[1]);
strcpy(out_str[(*pos)++], name_key[0]);
}
tmp = num % 100;
if(tmp == 0)
{
(*pos)--;
return ;
}
tmp = num % 100;
if((tmp / 10) == 1)
{
strcpy(out_str[(*pos)++], name_of_num[tmp]);
}
else if((tmp / 10) > 1)
{
strcpy(out_str[(*pos)++], name_of_numplus[tmp / 10]);
}
tmp = num % 10;
if(tmp != 0 && (((num % 100) / 10) != 1))
{
strcpy(out_str[(*pos)++], name_of_num[tmp % 10]);
}
}
int getName(int num, char (*out_str)[32], int *n)
{
if(num >= 1000000000)return -1;
int tmp;
*n = 0;
tmp = (num / 1000000) % 1000;
if(tmp > 0)
{
getNamePre(tmp, out_str, n);
strcpy(out_str[(*n)++], name_key[3]);
}
tmp = (num / 1000) % 1000;
if(tmp > 0)
{
getNamePre(tmp, out_str, n);
strcpy(out_str[(*n)++], name_key[2]);
}
getNamePre(num % 1000, out_str, n);
return 0;
}
void test(int num)
{
char buf[128][32] = {{0},};
int n;
getName(num, buf, &n);
int i;
for(i = 0; i < n; i++)
{
printf("%s ", buf[i]);
}
printf("\n");
}
int main(void)
{
int num;
while(scanf("%d", &num) != EOF)
test(num);
return 0;
}
8、迷宫问题
#include
#include
struct XYPOS{
int x;
int y;
};
void printPath(struct XYPOS *p, int n)
{
int i;
for(i = 0; i < n; i++)
{
printf("(%d,%d)\n", p[i].y, p[i].x);
}
//printf("\n");
}
struct XYPOS record[128] = {0};
int record_pos = 0;
struct XYPOS tmp_record[128] = {0};
int tmp_pos;
int getPath(int *data, int x_top, int y_top, struct XYPOS now)
{
int i, j;
struct XYPOS xypos;
int mark;
//记录
record[record_pos++] = now;
if((now.x >= 0 && now.x <= x_top) && (now.y >= 0 && now.y <= y_top))
{
for(i = 0; i < 4; i++)
{
xypos = now;
if(i == 0)
{
xypos.x++;
}
else if(i == 1)
{
xypos.x--;
}
else if(i == 2)
{
xypos.y++;
}
else if(i == 3)
{
xypos.y--;
}
//禁止环路
mark = 0;
for(j = 0; j < record_pos; j++)
{
if(record[j].x == xypos.x && record[j].y == xypos.y)
{
mark = 1;
break;
}
}
if(mark == 1)continue;
//越界
if(!(xypos.x >= 0 && xypos.x <= x_top) && (xypos.y >= 0 && xypos.y <= y_top))
{
continue;
}
//遇 1
if(data[xypos.y * (1 + x_top) + xypos.x] == 1)
{
continue;
}
//最终结果
if(xypos.x == x_top && xypos.y == y_top)
{
record[record_pos++] = xypos;
if(tmp_pos > record_pos)//取最短路径
{
tmp_pos = record_pos;
for(j = 0 ; j < record_pos; j++)
{
tmp_record[j] = record[j];
}
}
//printPath(record, record_pos);
record_pos--;
continue;
}
//递归
getPath(data, x_top, y_top, xypos);
}
}
record_pos--;
return 0;
}
int main(void)
{
int x, y, i, j;
int data[100] = {0};
while(scanf("%d %d", &y, &x) != EOF)
{
for(i = 0; i < y; i++)
{
for(j = 0; j < x; j++)
{
scanf("%d ",(data + i * x + j));
}
scanf("\n");
}
struct XYPOS now = {0, 0};
record_pos = 0;
tmp_pos = x*y;
getPath(data, x - 1, y - 1, now);
printPath(tmp_record, tmp_pos);
}
return 0;
}
9、数独问题
（部分用例通过 2/6,，通过率30%）
#include
#define TEST -1
int data[81] = {0};
int bit[81] = {0};
int stop = 0;
void printData(int *data)
{
int i;
for(i = 0; i < 81; i ++)
{
printf("%d ", *(data + i));
if((i + 1) % 9 == 0)
{
printf("\n");
}
}
}
int setnum(int n)
{
int i, j, x, y, mark, tmp_data, tmp_bit;
y = n / 9;
x = n % 9;
if(stop == 1)
{
return 0;
}
if(n >= 80)
{
//printf("OK\n");
printData(data);
//stop = 1;
return 0;
}
if(data[n] == 0)
{
if(n == TEST)printf("TEST = %d is coming.............[0]\n", TEST);
for(i = 0; i < 9; i++)
{
if(bit[y * 9 + i] == 0)
{
if(n == TEST)printf("==>[select] num=%d\n", i + 1);
for(j = 0 ; j < 9 ; j ++)
{
mark = 0;
if(data[x + j * 9] == (i + 1) && y != j)
{
if(n == TEST)printf("==>[not ok](%d,%d) data=%d\n", j, x, i + 1);
mark = 1;
break;
}
}
if(mark == 1)
continue;
if(n == TEST)printf("==>[ok] data=%d\n", i + 1);
tmp_data = data[n];
data[n] = i + 1;
tmp_bit = bit[y * 9 + i];
bit[y * 9 + i] = 1;
setnum(n + 1);
//回退
data[n] = tmp_data;
bit[y * 9 + i] = tmp_bit;
}
}
}
else
{
if(n == TEST)printf("TEST = %d is coming.............[1]\n", TEST);
setnum(n + 1);
}
return 0;
}
int main(void)
{
int i;
for(i = 0; i < 81; i ++)
bit[i] = 0;
for(i = 0; i < 81; i ++)
{
scanf("%d", data + i);
if((i + 1) % 9 == 0)
scanf("\n");
if(*(data + i) != 0)
bit[(i / 9) * 9 + (*(data + i) - 1)] = 1;
}
//printData(bit);
stop = 0;
setnum(0);
return 0;
}
10、漂亮数字
python 例子（写代码速度确实更快，用到了现成的“轮子”）
num = input()
while True:
try:
sum = 0
mystr = input().strip().lower()
str_len = len(mystr)
l = [];
value_a = ord('a')
for i in range(26):
tmp = mystr.count(chr(value_a + i))
if tmp != 0:
l.append(tmp)
l = sorted(l, reverse=True)
for i in range(len(l)):
sum += l[i] * (26 - i)
if(sum != 0):
print(sum)
except:
break
c语言
#include
#include
/*思路
*统计字符出现的次数，然后排序，次数大的给数值26，次之给25，依次类推，最后求和
*/
void sort(int *p, int n)//冒泡排序
{
int i, j;
for(i = 0; i < n -1; i++)
for(j = i + 1; j < n; j ++)
if(p[i] < p[j])
{p[i] ^= p[j];p[j] ^= p[i]; p[i] ^= p[j];}
}
int main(void)
{
int num, i, j, len, cnt, sum;
int bit[26] = {0};
int array[26] = {0};
char buf[4096] = {0};
while(scanf("%d", &num) != EOF)
{
for(i = 0; i < num; i++)
{
scanf("%s\n", buf);
len = strlen(buf);
for(j = 0; j < 26; j++)
bit[j] = 0;
for(j = 0; j < len; j++)
{
if(buf[j] >= 'a' && buf[j] <= 'z')
bit[buf[j] - 'a']++;
else if(buf[j] >= 'A' && buf[j] <= 'Z')
bit[buf[j] - 'A']++;
}
cnt = 0;
for(j = 0; j < 26; j++)
if(bit[j] != 0)
array[cnt++] = bit[j];
if(cnt == 0)
continue;
sort(array, cnt);
sum = 0;
for(j = 0; j < cnt; j++)
sum += array[j] * (26 - j);
printf("%d\n", sum);
}
}
return 0;
}
11、链表
#include
#include
struct myque{
int val;
struct myque *next;
};
#define NEW_NODE (struct myque *)malloc(sizeof(struct myque));
#define NULL_THEN_RETURN do{if(head == NULL) return;}while(0);
void que_print(struct myque *head)
{
NULL_THEN_RETURN;
struct myque *prob = head;
while(prob != NULL)
{
printf("%d ", prob->val);
prob = prob->next;
}
printf("\n");
}
struct myque * que_init(int val)
{
struct myque * q = NEW_NODE;
q->val = val;
q->next = NULL;
return q;
}
void que_insert(struct myque *head, int pos, int val)
{
NULL_THEN_RETURN;
struct myque *prob = head;
struct myque *tmp;
if(pos == val)
return;
while(prob != NULL)
{
if(prob->val == pos)
{
tmp = prob->next;
prob->next = NEW_NODE;
prob->next->val = val;
prob->next->next = tmp;
break;
}
prob = prob->next;
}
}
void que_del(struct myque *head, int val)
{
NULL_THEN_RETURN;
struct myque *prob = head;
struct myque *last = head;
struct myque *tmp;
while(prob != NULL)
{
if(prob->val == val)
{
if(prob == head)
{
tmp = head->next;
*head = *(head->next);
free(tmp);
return;
}
last->next = prob->next;
free(prob);
return;
}
last = prob;
prob = prob->next;
}
}
int main(void)
{
int num, head_v, i, val, pos, del_v;
while(scanf("%d", &num) != EOF)
{
scanf("%d", &head_v);
struct myque *head = que_init(head_v);
for(i = 0; i < num - 1; i++)
{
scanf("%d %d", &val, &pos);
que_insert(head, pos, val);
}
scanf("%d\n", &del_v);
que_del(head, del_v);
que_print(head);
}
return 0;
}