http://ac.jobdu.com/problem.php?pid=1021
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- 题目描述:
- 统计一个给定字符串中指定的字符出现的次数。
- 输入:
- 测试输入包含若干测试用例,每个测试用例包含2行,第1行为一个长度不超过5的字符串,第2行为一个长度不超过80的字符串。注意这里的字符串包含空格,即空格也可能是要求被统计的字符之一。当读到'#'时输入结束,相应的结果不要输出。
- 输出:
- 对每个测试用例,统计第1行中字符串的每个字符在第2行字符串中出现的次数,按如下格式输出:
c0 n0
c1 n1
c2 n2
...
其中ci是第1行中第i个字符,ni是ci出现的次数。
- 样例输入:
ITHIS IS A TESTi ngthis is a long test string#
- 样例输出:
I 2i 3 5n 2g 2
- #include<iostream>
- #include<cstdio>
- #include<cstdlib>
- #include<memory.h>
- #include<map>
- #define MAX 0xFFFFFFF
- using namespace std;
- map<char, int> wordmap;
- map<char, int>::iterator it;
- int main(){
- //freopen("in.txt", "r", stdin);
- string s1;
- string s2;
- while(getline(cin, s1), s1!="#"){
- int len1 = s1.length();
- for(int i=0;i<len1;++i){
- wordmap[s1[i]] = 0;
- }
- getline(cin, s2);
- int len2 = s2.length();
- for(int i=0;i<len2;++i){
- it = wordmap.find(s2[i]);
- if(it!=wordmap.end()){
- wordmap[s2[i]] = it->second + 1;
- }
- }
- for(int i=0;i<len1;++i){
- it = wordmap.find(s1[i]);
- printf("%c %d\n", s1[i], it->second);
- }
- wordmap.clear();
- }
- //fclose(stdin);
- }
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