题目
Eva would like to make a string of beads with her favorite colors so she went to a small shop to buy some beads. There were many colorful strings of beads. However the owner of the shop would only sell the strings in whole pieces. Hence Eva must check whether a string in the shop contains all the beads she needs. She now comes to you for help: if the answer is Yes, please tell her the number of extra beads she has to buy; or if the answer is No, please tell her the number of beads missing from the string.
For the sake of simplicity, let’s use the characters in the ranges [0-9], [a-z], and [A-Z] to represent the colors. For example, the 3rd string in Figure 1 is the one that Eva would like to make. Then the 1st string is okay since it contains all the necessary beads with 8 extra ones; yet the 2nd one is not since there is no black bead and one less red bead.
Figure 1
Input Specification:
Each input file contains one test case. Each case gives in two lines the strings of no more than 1000 beads which belong to the shop owner and Eva, respectively.
Output Specification:
For each test case, print your answer in one line. If the answer is Yes, then also output the number of extra beads Eva has to buy; or if the answer is No, then also output the number of beads missing from the string. There must be exactly 1 space between the answer and the number.
Sample Input 1:
ppRYYGrrYBR2258
YrR8RrY
Sample Output 1:
Yes 8
Sample Input 2:
ppRYYGrrYB225
YrR8RrY
Sample Output 2:
No 2
思路
用vector模拟map统计两个字符串中每个字符出现次数，统计第一个字符串与第二个字符串中元素差值。若差值>0，则输出No + 差值；否则输出Yes + 两字符串长度差。
一开始审题不够细，把题目想复杂了，以为商店的项链是可以截开的，当能满足需求时，要求出需要额外买的最小差值。于是分别从前端和后端不断截掉一个（计数–），没去掉一个计算一次是否仍然满足…
其实商店的项链是整根卖的，超出的数量直接等于两个字符串长度差。
代码
#include <iostream>
#include <vector>
using namespace std; int main() { string s1, s2;cin >> s1 >> s2;vector<int> m1(128), m2(128);for (int i=0; i<s1.size(); i++){m1[s1[i]]++;}for (int i=0; i<s2.size(); i++){m2[s2[i]]++;}int dif = 0;for (int i=0; i<128; i++){if (m1[i] < m2[i]){dif += (m2[i] - m1[i]);}}if (dif > 0){cout << "No " << dif << endl;}else{cout << "Yes " << s1.size()-s2.size() << endl;}return 0;
}