Educational Codeforces Round 82 (Rated for Div. 2)：A. Erasing Zeroes

太过爱你忘了你带给我的痛 2023-07-05 05:30 6阅读 0赞

**Discription**  
You are given a string ss. Each character is either 0 or 1.You want all 1’s in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1’s form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.You may erase some (possibly none) 0’s from the string. What is the minimum number of 0’s that you have to erase?

**Input**  
The first line contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.Then tt lines follow, each representing a test case. Each line contains one string ss (1≤|s|≤1001≤|s|≤100); each character of ss is either 0 or 1.

**Output**  
Print tt integers, where the ii-th integer is the answer to the ii-th testcase (the minimum number of 0’s that you have to erase from ss).

**Example**  
**input**  
3  
010011  
0  
1111000  
**output**  
2  
0  
0

**Note**  
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).

**题意**  
题意很简单就是给定一串用0和1组成的字符串，要求减去最少的0，让字符串中所有的1都在一块。

**思路**  
记录每个1所在的位置，将所有的差值减一都加起来就是答案。  
因为开始要判断字符串，被卡了一下。

**AC代码**

#include<bits/stdc++.h>
    using namespace std;
    int t;
    char s[110];
    int ans;
    vector<int>v;
    int main()
    { 
        scanf("%d",&t);
        while(t--)
        { 
            scanf("%s",s);
            ans=0;
            v.clear();
            for(int i=0; i<strlen(s); i++)
                if(s[i]=='1')
                    v.push_back(i);
            if(v.size()<2)
                ans=0;
            else
                for(int i=1; i<v.size(); i++)
                    ans+=(v[i]-v[i-1]-1);
            cout<<ans<<endl;
        }
        return 0;
    }

大佬的简单代码：

#include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    main()
    { 
        ll t;
        cin>>t;
        while(t--)
        { 
            ll n,a,b;
            cin>>n>>a>>b;
            ll m=(n+1)/2;
            cout<<m+max(n-m,(m-1)/a*b)<<endl;
        }
    }