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Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 650 Accepted Submission(s): 333
Author
UESTC

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w",stdout);
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int,int> PII;
const int MAXN = 10000005;
char s[MAXN];
int check(char s[], int mod){
//将数字看成多项式的和 比如12345就是1*10^5+2*10^4+3*10^3+4*10^2+5*10^1
//先把数求模和分开求再合起来求模结果是一样的
LL sum = 0;
int len = strlen(s);
for( int i = 0; i < len ; i++){
sum = sum*10 + s[i]-48;
sum %= mod;
}
return sum;
}
int main()
{
//FIN
int cas = 1;
while(~scanf("%s", s)){
if(check(s, 137) == 0 && check(s, 73) == 0) printf("Case #%d: YES\n",cas++);
else printf("Case #%d: NO\n",cas++);
}
}
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原文地址:http://www.cnblogs.com/Hyouka/p/5774231.html