HDU 5832 A water problem （水题，大数）

题意：给定一个大数，问你取模73 和 137是不是都是0.

析：没什么可说的，先用char 存储下来，再一位一位的算就好了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 100000000000000000;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e7 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
inline LL Max(LL a, LL b){  return a < b ? b : a; }
inline LL Min(LL a, LL b){  return a > b ? b : a; }
char s[maxn];
 
int main(){
    int kase = 0;
    while(scanf("%s", s) == 1){
        int ans1 = 0, ans2 = 0;
        n = strlen(s);
        for(int i = 0; i < n; ++i){
            ans1 = (ans1 * 10 + s[i] - '0') % 73;
            ans2 = (ans2 * 10 + s[i] - '0') % 137;
        }
 
        printf("Case #%d: %s\n", ++kase, !ans1 && !ans2 ? "YES" : "NO");
    }
    return 0;
}