2 条题解
-
0
#include<iostream> bool j(int y) { return (y % 4 == 0 && y % 100 != 0) || y % 400 == 0; } void solve(int date[3]) { int they{date[0]}, them{date[1]}, tdby{date[2]}; int m[13] = {29, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; if(tdby - 2 <= 0) { if(j(they) && them == 3) { tdby += m[0] - 2; them = 2; } else if(them == 1) { they--; them = 12; tdby += m[12] - 2; } else { them--; tdby += m[them] - 2; } } else { tdby -= 2; } printf("%d-%02d-%02d", they, them, tdby); } int main() { int date[3]; scanf( "%d-%d-%d", &date[0], &date[1], &date[2] ); solve(date); return 0; }
信息
- ID
- 45
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 9
- 标签
- 递交数
- 20
- 已通过
- 2
- 上传者