UVa 12210 - A Match Making Problem
来源:互联网 发布:广告机改网络电视 编辑:程序博客网 时间:2024/05/16 17:55
題目:單身那時和女士配對,每次最年長的男士和年齡最相近的女士配對,
如果男士都有伴侶則輸出0,否則輸出沒有伴侶的男士個數,并輸出年齡最小的男士年齡。
分析:簡單題。比較數量,輸出最小值即可。
如果,單身男士的數量不多餘單身女士的數量則輸出0;
否則,沒有伴侶的男士個數即為女士數量減去男士數量,并輸出年齡最小的男士年齡。
說明:這裡使用排序找最小值,枚舉更快,╮(╯▽╰)╭。
#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;int bachelor[10001];int spinster[10001]; int main(){int bachelor_size, spinster_size, cases = 0;while (~scanf("%d%d",&bachelor_size,&spinster_size)) {if (!bachelor_size && !spinster_size) {break;}for (int i = 0; i < bachelor_size; ++ i) {scanf("%d",&bachelor[i]);}sort(bachelor, bachelor+bachelor_size);for (int i = 0; i < spinster_size; ++ i) {scanf("%d",&spinster[i]);}printf("Case %d:",++ cases);if (bachelor_size <= spinster_size) {printf(" %d\n",0);}else {printf(" %d %d\n",bachelor_size-spinster_size,bachelor[0]);}}return 0;}
0 0
- UVa 12210 - A Match Making Problem
- UVa12210 - A Match Making Problem(排序)
- UVA_12210_A Match Making Problem
- A Dicey Problem UVA
- A Dicey Problem UVA
- A Research Problem UVA
- A Scheduling Problem UVA
- A Research Problem UVA
- Need help on a fuzzy match problem
- uva 387 - A Puzzling Problem
- UVA 11069 - A Graph Problem
- UVA 11069 A Graph Problem
- UVA 11069 A Graph Problem
- UVA 11021 Problem A Tribbles
- uva 387 A Puzzling Problem
- UVA 387 - A Puzzling Problem
- UVa 11069 - A Graph Problem
- UVA - 387 A Puzzling Problem
- 6. ZigZag Conversion
- android四大组件--ContentProvider详解
- hduoj1115
- Javo 基础 流的分类
- linux 添加一个系统调用
- UVa 12210 - A Match Making Problem
- R tutorial 02 - Operators 运算元
- python django paginator分页
- String、StringBuffer与StringBuilder之间区别
- 简图记录-程序的生命周期:编译链接装载运行
- Asp.net MVC中Html.Partial, RenderPartial, Action,RenderAction 区别和用法
- HTML5 新元素
- R tutorial 03 - Function 函数
- 深入理解数据库磁盘存储(Disk Storage)