ZOJ 3326 - An Awful Problem
来源:互联网 发布:linux注释#作用 编辑:程序博客网 时间:2024/05/17 02:20
题意:给定一个日期范围,判断月和日均为素数的天数。
比较水的题目,for循环中的?:判断语句、日期的打表,可以大幅精简代码。
#include <iostream>#include <cstdio>using namespace std;int day1[15] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; //闰年int day2[15] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; //平年inline bool is_run(const int x) { return x%400==0 || (x%4==0 && x%100!=0);}bool is_prime(const int x) { if(x==0 || x==1) return false; for(int i = 2; i < x; ++i) if(x % i == 0) return false; return true;}int main() { int T; cin >> T; while(T--){ int y1, y2, m1, m2, d1, d2; cin >> y1 >> m1 >> d1; cin >> y2 >> m2 >> d2; int cnt = 0; for(int y = y1; y <= y2; ++y){ for(int m = (y==y1 ? m1 : 1); m <= (y==y2 ? m2 : 12); ++m){ for(int d = (y==y1&&m==m1 ? d1 : 1); d <= (y==y2&&m==m2 ? d2 : (is_run(y) ? day1[m] : day2[m])); ++d){ if(is_prime(m) && is_prime(d)) ++cnt; } } } cout << cnt << endl; } return 0;}
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