南邮2014程序设计新生赛暨蓝桥杯校内自主选拔赛
南邮2014程序设计新生赛暨蓝桥杯校内自主选拔赛
题目链接: http://acm.njupt.edu.cn/acmhome/contest.do?&method=contestDetail&contestId=213
好奇的奥利安娜
时间限制(普通/Java):1000MS/3000MS 运行内存限制:65536KByte
总提交:333 测试通过:99
总提交:333 测试通过:99
描述
输入
输出
样例输入
样例输出
呵呵
#include <cstdio>
int main()
{
int n,a,b,c,d;
scanf("%d", &n);
while(n--)
{
scanf("%d%d%d%d", &a, &b, &c, &d);
if(a + b + c + d == 0)
printf("sidajiekong\n");
else
printf("%d\n", a+b+c+d);
}
}
突发奇想的茂凯
Time Limit(Common/Java):1000MS/3000MS Memory Limit:65536KByte
Total Submit:106 Accepted:26
Total Submit:106 Accepted:26
Description
Input
Output
Sample Input
Sample Output
12-1
继续呵呵
#include <cstdio>
#include <algorithm>
using namespace std;
int a[21];
bool cmp(int a, int b)
{
return a > b;
}
bool judge(int a, int b, int c)
{
if(a + b > c && a + c > b && b + c > a)
return true;
return false;
}
int main()
{
int T, ma;
scanf("%d", &T);
while(T--)
{
int n;
ma = -1;
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
sort(a, a + n, cmp);
for(int i = n - 1; i >= 0; i--)
for(int j = i - 1; j >= 0; j--)
for(int k = j - 1; k >= 0; k--)
if(judge(a[i], a[j], a[k]))
ma = max(ma, a[i] + a[j] + a[k]);
printf("%d\n", ma);
}
}
马尔扎哈的疑惑
Time Limit(Common/Java):1000MS/3000MS Memory Limit:65536KByte
Total Submit:160 Accepted:27
Total Submit:160 Accepted:27
Description
Input
Output
Sample Input
Sample Output
46
呵呵呵
#include <cstdio>
int a[60];
int main()
{
int n;
scanf("%d", &n);
int x;
a[1] = 1;
a[2] = 2;
a[3] = 3;
for(int i = 4; i <= 100; i++)
a[i] = a[i - 1] + a[i - 3];
while(n--)
{
scanf("%d", &x);
printf("%d\n",a[x]);
}
}
大发明家的童年
Time Limit(Common/Java):1000MS/3000MS Memory Limit:65536KByte
Total Submit:28 Accepted:4
Description
Input
Output
Sample Input
Sample Output
-1 0 0 0 -V +1 0 0 0 +V -1 +1 -B 0 0 -V +1 -1小模拟，暴力模拟
#include <cstdio>
char str[205];
int main()
{
int n;
scanf("%d", &n);
getchar();
for(int i = 0; i < 2 * n - 1; i++)
scanf("%c ", &str[i]);
scanf("%c", &str[2 * n - 1]);
for(int i = 0; i < n; i ++)
if(str[i] == '0' && str[i + 1] == '0' && str[i + 2] == '0' && str[i + 3] == '0')
str[i + 3] = 'V';
int cnt;
for(int i = 0 ; i < n; i++)
{
cnt = 0;
for(int j = i + 1; j < n; j++)
{
if(str[i] == 'V' && str[j] == 'V')
{
for(int k = i; k <= j; k++)
if(str[k] == '1')
cnt++;
if(cnt % 2 == 0)
{
str[j - 1] = str[j - 2] = '0';
str[j - 3] = 'B';
}
break;
}
}
}
bool flag = true;
for(int i = 0; i < n - 1; i++)
{
if(str[i] == '0')
printf("%c ", str[i]);
else
{
if(str[i] == '1' && flag)
{
printf("-1 ");
flag = !flag;
}
else if(str[i] == '1' && !flag)
{
printf("+1 ");
flag = !flag;
}
else if(str[i] == 'V' && !flag)
printf("-V ");
else if(str[i] == 'V' && flag)
printf("+V ");
else if(str[i] == 'B' && !flag)
{
printf("+B ");
flag = !flag;
}
else if(str[i] == 'B' && flag)
{
printf("-B ");
flag = !flag;
}
}
}
if(str[n-1] == '0')
printf("0\n");
else if(str[n-1] == '1' && flag)
printf("-1\n");
else if(str[n-1] == '1' && !flag)
printf("+1\n");
else if(str[n-1] == 'V' && !flag)
printf("-V\n");
else if(str[n-1] == 'V' && flag)
printf("+V\n");
}
爱魔法的露露
Time Limit(Common/Java):1200MS/4000MS Memory Limit:65536KByte
Total Submit:55 Accepted:7
Total Submit:55 Accepted:7
Description
Input
Output
Sample Input
Sample Output
Hint
-
读入数据请使用 scanf()；
-
对于样例，共有{1},{2},{3},{1,2},{2,3},{1,2,3}，6个魔法区间，所以威力为6。
hash ＋ 线性两点扫
#include <cstdio>
#include <cstring>
int const MAX = 1e6 + 5;
int a[MAX], hash[MAX];
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int n;
scanf("%d", &n);
memset(hash, 0, sizeof(hash));
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int ed = 1, st = 1;
int ans = 0;
for( ; ed <= n; ed++)
{
if(!hash[a[ed]])
hash[a[ed]] ++;
else
break;
}
while(st <= n)
{
hash[a[st]] --;
ans += ed - st;
st++;
for( ; ed <= n; ed++)
{
if(!hash[a[ed]])
hash[a[ed]] ++;
else
break;
}
}
printf("%d\n", ans);
}
}
赵信的往事
Time Limit(Common/Java):1000MS/3000MS Memory Limit:65536KByte
Total Submit:29 Accepted:9
Total Submit:29 Accepted:9
Description
Input
Output
Sample Input
Sample Output
Hint
对于样例一，6的素因子集合为{2,3},与其他人不同，所以不行；
对于样例二，所有数的素因子集合均为{3},因此可以组团。
无限gcd
#include <cstdio>
#include <algorithm>
using namespace std;
int a[5], flag;
int gcd(int a, int b)
{
return b ? gcd(b, a % b) : a;
}
void cal(int x,int y)
{
int tmp;
if(x == 1 && y == 1)
return;
tmp = gcd(x, y);
x /= tmp;
y /= tmp;
if(x == 1 && y == 1)
return;
else if(tmp == 1 && y % x != 0)
{
flag = 0;
return;
}
else
cal(x, g);
return;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
flag = 1;
scanf("%d %d %d", &a[0], &a[1], &a[2]);
sort(a,a+3);
cal(a[0],a[1]);
cal(a[1],a[0]);
if(flag)
{
cal(a[0],a[2]);
cal(a[2],a[0]);
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
}