ACM_Alien And Password

Alien And Password

Time Limit: 2000/1000ms (Java/Others)

Problem Description:

Alien Fred wants to destroy the Earth, but he forgot the password that activates the planet destroyer.
You are given a string S. Fred remembers that the correct password can be obtained from S by erasing exactly one character.
Write a program to calculate the number of different passwords Fred needs to try.
0)	 	
"A"
Answer: 1
In this case, the only password Fred needs to try is an empty string.
1)   	
"ABA"
Answer: 3
The following three passwords are possible in this case: "BA", "AA", "AB".
2)	  	
"AABACCCCABAA"
Answer: 7
3)    	
"ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ"
Answer: 1
Regardless of which character we erase, we will always obtain the same string. Thus there is only one possible password: the string that consists of 49 'Z's.

Input:

The input contains multiple cases.Each case contains a string(length less than 100).

Output:

For each case,output the answer of the problem.

Sample Input:

A
ABA
AABACCCCABAA
ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

Sample Output:

1
3
7
1
解题思路：使用string.erase(pos,num)，删除从pos索引开始的num个字符, 返回*this，为了不改变目标字符串str，所以用临时的字符串obj保存删除str中每个位置上的字符前的字符串str，这样每次删除后都将其放在容器set中，最后输出容器中元素的个数即可，水过！
AC代码：

#include<bits/stdc++.h>
using namespace std;
int main(){
    string str,obj;set<string> st;
    while(cin>>str){
        st.clear();//清空
        for(size_t i=0;i<str.length();++i){
            obj=str;st.insert(obj.erase(i,1));
        }
        cout<<st.size()<<endl;
    }
    return 0;
}