HDU - 4768 Flyer

来源：互联网发布：淘宝怎么做全球购编辑：程序博客网时间：2024/05/12 12:52

题意：很好理解就不说了，用了异或，如果出现偶数次就会等于0，所以最后的数就是答案了，然后再根据num=A[i]+k*c[i],所以减去A[i]就是c[i]的倍数了，这就是sum++的判断条件，被杭电的输入，输出坑了很久

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int N = 20010;int n;long long a[N], b[N], c[N];int main(){    while(~scanf("%d", &n)){        long long num = 0;        for(int i = 1; i <= n; i++){            cin>>a[i]>>b[i]>>c[i];            for(long long j = a[i]; j <= b[i]; j+=c[i]){                num ^= j;            }        }        long long sum = 0;        for(int i = 1; i <= n; i++){            if((num-a[i])%c[i]==0 && (num>=a[i])&&(num<=b[i]))                sum++;        }        if(num==0){            puts("DC Qiang is unhappy.");            continue;        }        cout << num << " " << sum << endl;    }    return 0;}