uva10474
来源:互联网 发布:网络技术交流论坛 编辑:程序博客网 时间:2024/05/17 04:33
输出x在A中排第几小或输出没有x.
#include<cstdio>#include<algorithm>using namespace std;int a[100000];int main() {int m, n; int x; int k = 0;while (scanf_s("%d%d",&m,&n)==2&&m) {printf("CASE# %d:\n", ++k);for (int i = 0; i < m; i++) {scanf_s("%d",&a[i]);}sort(a, a + m);for (int i = 0; i < n; i++) {scanf_s("%d", &x);int q = lower_bound(a, a + m, x) - a;//注意这个函数的用法if (q < m&&a[q] == x)printf("%d found at %d\n", x, q+1);elseprintf("%d not found\n", x);}}return 0;}
0 0
- uva10474
- UVA10474
- uva10474
- Uva10474
- UVA10474
- uva10474
- UVA10474
- Uva10474
- UVA10474
- uva10474
- Uva10474
- sort--uva10474
- UVA10474-弹珠在哪里
- e5-1 uva10474
- uva10474 (二分查找)
- UVa10474(排序和查找)
- UVA10474查找优化
- uva10474 - Where is the Marble?
- 电脑硬盘修复方法
- String hdoj 5672(字符串追赶)
- PAT (Advanced Level) Practise 1107. Social Clusters (30) 并查集
- ASP.NET MVC Code First要点
- C++高斯赛德迭代法,求线性方程组的解(version1.0)
- uva10474
- 【黑苹果】Mac下修复iMessage,FaceTime等无法登陆
- unity delegate事件委托
- Thrift之Protocol源码分析
- 287. Find the Duplicate Number
- java包装类的小结
- 专题二 Problem1011
- 基于Windows 2012配置SQL Server 2014 AlwaysOn
- 小白说编译原理-4-计算器yacc