#297 (div.2) A. Vitaliy and Pie
来源:互联网 发布:淘宝参加聚划算的要求 编辑:程序博客网 时间:2024/06/05 17:18
1.题目描述:点击打开链接
2.解题思路:本题其实就是一道模拟题,每次遇到小写字符,就把对应的个数加一,遇到大写字符,如果有钥匙,个数减一,否则cnt++。输出cnt即可。但比赛时候用了过多的STL,导致结果TLE了,导致我第一次A题都跪了==。其实直接用普通的标记数组就可以统计个数了。以后凡是能够用到C的,应该尽量用C,保证简洁与高效。
3.代码:
#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;#define N 1000000+10int vis[26];int main(){//freopen("t.txt", "r", stdin);int n;char s[N];while (~scanf("%d", &n)){memset(vis, 0, sizeof(vis));scanf("%s", s); int len = 2 * n - 2;int cnt = 0;for (int i = 0; i < len;i++)if (i & 1){if (vis[s[i] - 'A'])vis[s[i] - 'A']--;else cnt++;}else vis[s[i] - 'a']++;cout << cnt << endl;}return 0;}
0 0
- #297 (div.2) A. Vitaliy and Pie
- Codeforces #297 (Div. 2) A. Vitaliy and Pie (水
- A. Vitaliy and Pie
- A. Vitaliy and Pie
- A. Vitaliy and Pie
- Codeforces Round #297 (Div. 2) —— A. Vitaliy and Pie
- A. Vitaliy and Pie(Codeforces Round #297 (Div. 2) 水题)
- Codeforces Round #297 (Div. 2) A - Vitaliy and Pie 解题报告
- Codeforces Round#297 A.Vitaliy and Pie
- CF 525A(Vitaliy and Pie-模拟)
- coderforce 525A Vitaliy and Pie
- Codeforces_round297_A.Vitaliy and Pie
- Vitaliy and Pie
- CodeForces 525A Vitaliy and Pie 题意题
- #281 (div.2) A.Vasya and football
- #299 (div.2) A. Tavas and Nafas
- #308 (div.2) A. Vanya and Table
- #309 (div.2) A. Kyoya and Photobooks
- The Compiler (关于编译的点点滴滴)
- 看看老外的开源项目
- OJ 系列之分解字符串
- maven ArtifactTransferException
- Mysql配置与使用
- #297 (div.2) A. Vitaliy and Pie
- pdf转word免费软件
- 学习jQuery的on事件
- 02-线性结构1. Reversing Linked List (25)
- HEX,BYTE,ASC,CHAR,BCD互转
- 转:java日志记录最佳实践
- 导入导出thunderbird(雷鸟)中的邮件
- Android 如何监听返回键,弹出一个退出对话框
- 看到的不错的产品助理面试题