@[TOC]
Contest100000575 - 《算法笔记》3.1小节——入门模拟->简单模拟
1814 Problem A 剩下的树
来自 http://codeup.cn/contest.php?cid=100000575
#include
#include
using namespace std;
int tree[10005] = {0};
int main()
{
int L,M;
while(scanf("%d%d",&L, &M) != EOF)
{
if(L==0&&M==0)
return 0;
else
{
for(int i=0;i<=L;i++)
{
tree[i] = 1;
}
int sum = L+1;
while(M--)
{
int left,right;
scanf("%d%d",&left, &right);
for(int i=left;i<=right;i++)
{
tree[i] = 0;
}
}
for(int j=0;j<=L;j++)
{
if(tree[j] == 0)
sum--;
}
printf("%d\n",sum);
}
}
return 0;
}
1817 Problem B A+B
来自 http://codeup.cn/contest.php?cid=100000575
#include
#include
#include
using namespace std;
char A[15],B[15];
long trans(char *arr,int length)
{
long num=0,j=1;
for(int i=length-1;i>=0;i--)
{
if(arr[i]>='0' && arr[i]<='9')
{
num = num + (arr[i] - '0')*j;
j*=10;
}
}
if(arr[0]=='-')
num = -num;
return num;
}
int main()
{
long a,b;
while(scanf("%s%s",A, B) != EOF)
{
int lenA = strlen(A);
int lenB = strlen(B);
a = trans(A,lenA);
b = trans(B,lenB);
printf("%ld\n",a+b);
}
return 0;
}
1906 Problem C 特殊乘法
来自 http://codeup.cn/contest.php?cid=100000575
#include
#include
#include
using namespace std;
char a[15],b[15];
int num1[15],num2[15];
void trans(char *arr,int *num)
{
int len = strlen(arr);
for(int i=0;i 2036 Problem D 比较奇偶数个数
来自 http://codeup.cn/contest.php?cid=100000575
#include
#include
#include
using namespace std;
int main()
{
int n;
while(scanf("%d",&n) != EOF)
{
int num;
int countodd=0;
int counteven=0;
for(int i=0;i countodd)
{
printf("%s\n","NO");
}
else
printf("%s\n","YES");
}
return 0;
}
6116 Problem E Shortest Distance (20)
来自 http://codeup.cn/contest.php?cid=100000575
6128 Problem F A+B和C (15)
来自 http://codeup.cn/contest.php?cid=100000575
#include
#include
#include
using namespace std;
int main()
{
int T;
scanf("%d",&T);
int num=1;
while(T--)
{
long long a,b,c;
scanf("%ld%ld%ld",&a, &b, &c);
if(a + b > c)
{
printf("Case #%d: true\n",num);
}
else
{
printf("Case #%d: false\n",num);
}
num++;
}
return 0;
}
6129 Problem G 数字分类 (20)
来自 http://codeup.cn/contest.php?cid=100000575
#include
#include
#include
#include
using namespace std;
int A[5][1005];
int num[1005];
int main()
{
int N;
while(scanf("%d",&N) != EOF)
{
int A1_even=0,A2_jiaocuo=0,A3_count=0,A4_count=0,A5_max=0;
// int A3_count = 0,A4_count = 0;
float A4_avg = 0;
float A4_sum = 0;
int flag=0;
int A2_exist = 0;
for(int i=0;i>num;
switch(num%5)//switch语句正合适
{
case 0:
if(num % 2==0)
{
A1_even += num;
}
break;
case 1:
A2_jiaocuo+=(num * pow((double)(-1),flag));
flag++;
A2_exist = 1;
break;
case 2:
A3_count++;
break;
case 3:
A4_count++;
A4_sum += num;
break;
case 4:
if(A5_max < num)
A5_max = num;
break;
default:
break;
}
// cout< 6170 Problem H 部分A+B (15)
来自 链接: [link] http://codeup.cn/contest.php?cid=100000575
#include
#include
#include
using namespace std;
char A[15],B[15];
//法一竟然GG，不晓得要闹哪样
/*
int main()
{
char a,b;
while(scanf("%s%s%s%s",&A, &a, &B, &b) != EOF)
{
int lena = strlen(A);
int lenb = strlen(B);
int numa=0,numb=0;
int weight = 1,weightb = 1;
for(int i=0;i 6172 Problem I 锤子剪刀布 (20)
来自 http://codeup.cn/contest.php?cid=100000575
//题目较为繁琐，根据题意模拟
#include
#include
#include
/*
思路：每次输入进行比较。甲负的次数就是乙赢的次数，不用额外记录。最后输出甲乙获胜最多的手势，因为要考虑解不唯一，所以我采用把结果枚举。按字典序，J次数必须大于B和C，C次数必须大于B，可以大于等于B，B大于等于B、J就行。
注意：scanf会把'\n'读入，所以可能输入五组数据，就跳出结果了，要用getchar()来吸收。另外，判断要用if-else，不能用多个if，而没有else，这样会记录次数出现错误。
---------------------
作者：Wonz5130
来源：CSDN
原文：https://blog.csdn.net/Wonz5130/article/details/79844683
版权声明：本文为博主原创文章，转载请附上博文链接！
*/
using namespace std;
int main()
{
int jiasu=0,jiafa=0,jiaeq=0;
int jiac=0,jiaj=0,jiab=0,yic=0,yij=0,yib=0;
int N;
scanf("%d",&N);
while(N--)
{
getchar();
char a,b;
scanf("%c %c",&a,&b);
if(a == b)
jiaeq++;
//甲赢
else if(a == 'C' && b =='J')
{
jiasu++;
jiac++;
}
else if(a == 'J' && b=='B')
{
jiasu++;
jiaj++;
}
else if(a == 'B' && b=='C')
{
jiasu++;
jiab++;
}
//乙赢
else if(a == 'J' && b=='C')
{
jiafa++;
yic++;
}
else if(a == 'B' && b=='J')
{
jiafa++;
yij++;
}
else if(a == 'C' && b=='B')
{
jiafa++;
yib++;
}
// cout< jiab && jiaj > jiac)
{
printf("%c ",'J');
}
else if(jiac>jiab)
{
printf("%c ",'C');
}
else
{
printf("%c ",'B');
}
if(yij > yib && yij > yic)
{
printf("%c ",'J');
}
else if(yic>yib)
{
printf("%c ",'C');
}
else
{
printf("%c ",'B');
}
printf("\n");
return 0;
}
错误记录：
1.关于字符的控制台输入回车直接算一行了，截图就是本应输入10个数，没有getchar()只输入5个书
scanf("%d",&N);
getchar();
char a,b;
scanf("%c %c",&a,&b);
错误记录1
2.格式化输入输出没有搞清楚
printf("%d %d %d\n",&jiasu, &jiaeq, &jiafa);
printf("%d %d %d\n",&jiafa, &jiaeq, &jiasu);
出错截图：
出错截图2