HDU 5832A water problem 简单模拟

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看题可以容易知道给定的数得是10001的倍数
n 必须用字符串储存，然后模拟减去10001 再除10，继续重复这个步骤

#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <string>#include <vector>#include <map>#include <cstdio>#include <cstdlib>using namespace std;const int maxn = 10000000 + 7;char s[maxn];int a[maxn];void solve(){    int n = strlen(s);    for(int i = 0; i < n; i++)    {        a[i] = s[i] - '0';    }    for(int i = n - 1; i - 4 >= 0; i--)    {        int j = i - 4;        a[j] -= a[i];        a[i] -= a[i];        while(a[j] < 0 && j < n)        {            a[j] += 10;            a[j - 1]--;            j++;        }    }**HDU 5832 A water problem**    for(int i = 0; i < n; i++)    {        if(a[i] != 0)        {            cout << "NO" << endl;            return ;        }    }    cout << "YES" << endl;}int main(){    int cases = 0;    while(~scanf("%s", s))    {        printf("Case #%d: ", ++cases);        solve();        memset(a, 0, sizeof(a));        memset(s, 0, sizeof(s));    }    return 0;}

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