Codeforces Round #242 (Div. 2) A. Squats
来源:互联网 发布:车载导航wifi连接网络 编辑:程序博客网 时间:2024/05/09 20:26
Pasha has many hamsters and he makes them work out. Today, n hamsters (n is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.
For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?
The first line contains integer n (2 ≤ n ≤ 200;n is even). The next line contains n characters without spaces. These characters describe the hamsters' position: thei-th character equals 'X', if thei-th hamster in the row is standing, and 'x', if he is sitting.
In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.
4xxXx
1XxXx
2XX
1xX
6xXXxXx
0xXXxXx
问最少多少步,使得X与x数量相同并输出
#include<iostream>#include<cstring>#include<cstdio>#include<queue>#include<cmath>#include<algorithm>#define LL __int64#define inf 0x3f3f3f3fusing namespace std;char now[10];char s[210];bool bj;int main(){ int n,m,i,j,k; while(scanf("%d",&n)!=EOF) { bj=false; getchar(); scanf("%s",s); k=strlen(s); int l=k; j=0; for(i=0; i<k; i++) { if(s[i]=='X') j++; } if(n/2==j) { printf("0\n"); printf("%s\n",s); } else if(n/2<j) { printf("%d\n",j-n/2); int sum=0; for(i=0;i<l;i++) { if(sum==(j-n/2)) break; if(s[i]=='X') { s[i]='x'; sum++; } } printf("%s\n",s); } else { printf("%d\n",n/2-j); int sum=0; for(i=0;i<l;i++) { if(sum==(n/2-j)) break; if(s[i]=='x') { s[i]='X'; sum++; } } printf("%s\n",s); } } return 0;}
- Codeforces Round #242 (Div. 2) A. Squats
- Codeforces Round #242 (Div. 2) A. Squats
- Codeforces Round #242 (Div. 2) A. Squats
- codeforces 424A Squats
- Codeforces Round #242 (Div. 2) <A-D>
- Codeforces Round #242 (Div. 2) (A、B、C)
- Codeforces Round #131 (Div. 2) A B
- Codeforces Round #174 (Div. 2) Problem A
- Codeforces Round #173 (Div. 2) A.Bit++
- Codeforces Round #181 (Div. 2) A题
- Codeforces Round #184 (Div. 2) A题
- Codeforces Round #185 (Div. 2)--A,B
- Codeforces Round #172 (Div. 2) A题
- Codeforces Round #166 (Div. 2) A题
- Codeforces Round #132 (Div. 2) A题
- Codeforces Round #131 (Div. 2) A题
- Codeforces Round #133 (Div. 2) A题
- Codeforces Round #137 (Div. 2) A题
- UI计算器的设计
- nginx+FastCGI+c++
- 粗略。。Java项目设计模式之笔记----studying
- GDOI2016模拟8.19总结
- mysql5.6 ERROR 1045 (28000): Access denied for user (using password: NO)
- Codeforces Round #242 (Div. 2) A. Squats
- 绿色版mysql启动
- 机器学习那些事
- Solr 关于Analyzer、Tokenizer、和Filter,以及中文分词器
- Android设置虚线、圆角、渐变
- mongodb 时间格式转换和时区问题
- 程序中读取c语言中的一些问题
- 字符串常用函数
- android 获取网络连接(登陆验证+获取数据)