PAT basic 1057
来源:互联网 发布:诺亚方舟 知乎 编辑:程序博客网 时间:2024/05/29 09:37
#include <iostream>#include <cctype>#include <string>using namespace std;int main() { string s; getline(cin, s); int n = 0; for(int i = 0; i < s.length(); i++) { if(isalpha(s[i])) { s[i] = toupper(s[i]); n += (s[i] - 'A' + 1); } } int cnt0 = 0, cnt1 = 0; while(n != 0) { if(n % 2 == 0) { cnt0++; } else { cnt1++; } n = n / 2; } printf("%d %d", cnt0, cnt1); return 0;}/*分析:用getline接收一行字符串,对于字符串的每一位,如果是字母(isalpha),则将字母转化为大写,并累加(s[i] – ‘A’ + 1)算出n,然后将n转化为二进制,对每一位处理,如果是0则cnt0++,如果是1则cnt1++,最后输出cnt0和cnt1的值~~~*/
阅读全文
0 0
- PAT basic 1057
- PAT Basic
- PAT (Basic Level) Practise
- PAT Basic 1001
- PAT Basic 1002
- PAT Basic 1005
- PAT Basic 1006
- PAT Basic 1007
- PAT Basic 1008
- PAT Basic 1009
- PAT Basic 1010
- pat basic level 1016
- pat basic level 1018
- pat basic level 1019
- PAT(basic level)题解
- PAT basic 1004 : 成绩排名
- PAT (Basic) 1001~1005
- PAT (Basic) 1006~1010
- C++读入输出优化
- PAT basic 1056
- .net core 2.0学习笔记(四):迁移.net framework 工程到.net core
- poj 2112 Optimal Milking 最大流建模
- eclipse操作(二)
- PAT basic 1057
- Zemax中高斯光束设置的相关问题
- sqlmap教程2
- PAT basic 1058
- linux邮件系统
- 正则表达式-限定符_转义字符
- PAT basic 1059
- HDU-6154 CaoHaha's staff (找规律+二分)
- Centos7.2环境RPM 安装MySQL5.6.24