poj2846 The Bank of Kalii

来源：互联网发布：外汇期货实时数据接口编辑：程序博客网时间：2024/05/17 08:06

题意：给定前后日期，判断差距天数等。

思路：模拟

学习点：主要有三点：

(1) 调试程序时，重定向的使用

#define LOCAL //提交时删去。#include <iostream>int main(){ //新掌握1 !!#ifdef LOCALfreopen("2846.in", "r", stdin);freopen("2846.out", "w", stdout);#endif......}

(2)判断闰年

int isLeapYear(int year, int month){if (month > 2 && (year % 4 == 0) && (year % 100 || year % 400 == 0))return 1;return 0;}

(3)一种字符输出方法，比较好的技巧（见代码掌握点2）

printf("%d DAY%s PRIOR/n", -diff, diff == -1 ? "" : "S");

源代码：

//#define LOCAL#include <iostream>using namespace std;//新掌握3 !! int isLeapYear(int year, int month);int nMonthDay[] = {0,31, 28, 31, 30,31, 30, 31, 31,30, 31, 30, 31};int main(){ //新掌握1 !!#ifdef LOCALfreopen("2846.in", "r", stdin);freopen("2846.out", "w", stdout);#endifint nCase, nCaseNum = 0;;int M, D, Y, m, d, y;int day1, day2, diff;for (int i = 1; i <= 12; i ++){nMonthDay[i] += nMonthDay[i-1];}scanf("%d", &nCase);while (nCase --){scanf("%d/%d/%d %d/%d", &M, &D, &Y, &m, &d);if (M == m || M + 1 == m || M - 1 == m){y = Y;day1 = nMonthDay[M-1] + D + isLeapYear(Y, M);day2 = nMonthDay[m-1] + d + isLeapYear(y, m);diff = day2 - day1;}else if (M == 12 && m == 1){y = Y + 1;diff = d + (31 - D);}else if (M == 1 && m == 12){y = Y - 1;diff = -(D + 31 - d);}else{printf("%d OUT OF RANGE/n", ++nCaseNum);continue;}if (diff == 0){printf("%d SAME DAY/n", ++nCaseNum);}else if (diff < -7 || diff > 7){printf("%d OUT OF RANGE/n", ++nCaseNum);}else{printf("%d %d/%d/%d IS ", ++nCaseNum, m, d, y);if (diff < 0){//新掌握2 !!printf("%d DAY%s PRIOR/n", -diff, diff == -1 ? "" : "S");}else{printf("%d DAY%s AFTER/n", diff, diff == 1 ? "" : "S");}}}return 0;}int isLeapYear(int year, int month){if (month > 2 && (year % 4 == 0) && (year % 100 || year % 400 == 0))return 1;return 0;}