题目大意:给出两个时间,问从第一个时间变成第二个时间分针会和时针重叠几次。
解题思路:两个针重叠的时间是固定的,只要处理出这些重叠的时刻,在判断说给得时间区间包含的个数即可。
#include <cstdio> #include <cstring> #include <cmath> const int T = 12 * 60 * 100; const int D = 6545; int sh, sm, eh, em; int solve (int s, int t) { if (t < s) t += T; return t / D - (s - 100) / D; } int main () { printf("Program 3 by team X\n"); printf("Initial time Final time Passes\n"); while (scanf("%d%d%d%d", &sh, &sm, &eh, &em) == 4) { int s = (sh * 60 + sm) * 100; int t = (eh * 60 + em) * 100; printf(" %02d:%02d %02d:%02d", sh, sm, eh, em); printf("%8d\n", solve(s, t)); } printf("End of program 3 by team X\n"); return 0; }
uva 1529 - Clock(数论),码迷,mamicode.com
原文地址:http://blog.csdn.net/keshuai19940722/article/details/24738925