1092. To Buy or Not to Buy

题意： 

商家只卖第一串

买家是第二串

问买家需要的珠子在第一串中满不满足。若满足，看看多买了多少珠子。否则看看缺了哪些

思路：

太简单，哈希表都不需要用，直接用数组

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<stack>#include<vector>#include<queue>#include<string>#include<map>using namespace std;#define INF 99999999#define M 700int main(){string s1,s2;int hash[1000];int moren=0,lessn=0;memset(hash,0,sizeof(hash));int i,j;cin>>s1;for(i=0;i<s1.length();i++){hash[ s1[i] ]++;}cin>>s2;for(i=0;i<s2.length();i++){if(hash[ s2[i] ]==0){lessn++;}else hash[s2[i]]--;}if(lessn ==0){printf("Yes %d\n",s1.length() - s2.length());}elseprintf("No %d\n",lessn);return 0;}