D(1909)Perfect Chocolate

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Description

There is a chocolate, is composed of black and white small pieces. Xiao Ming is very fond of this chocolate, and the absolute difference between the number of black pieces and the number of white pieces is not more than 1, he think this chocolate is perfect.
Now Xiao Ming has one this chocolate in his hand, maybe it is not perfect.He will cut it into small blocks, making these small blocks are all perfect, he wanted to know how many times he will cut at least.

Input

There are multiple test cases.
For each test case.There is only one string composed of ‘0’ and ‘1’.’0’ is for the white piece while ‘1’ for the black piece.The length of the string for each case is not more than 100000.

Output

For each test case, you output one line “Case #%d:%d”

Sample Input

100111000100001

Sample Output

Case #1:3Case #2:0

Hint

Source

Author

HNU

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题意：当黑色小块与白色小块的个数相差的绝对值不大于1,就是一块Perfect Chocolate，给出的巧克力可能不是Perfect的，问最少切几下可以让所有的巧克力都是Perfect

思路：白的为-1，黑的为1，将待分割的这块巧克力最大化，找到能到达的最大值pos，并将pos+1作为下一块巧克力的起点

#include<iostream>#include<cmath>#include<string>using namespace std;int c[100005],sumb,sumw;int main(){    string str;int cas=0;    while(cin>>str){        int len=str.length();        sumb=sumw=0;        for(int i=0;i<len;i++){            if(str[i]=='1') sumb++,c[i]=1;            else            sumw++,c[i]=-1;        }        int ans=0;        cout<<"Case #"<<++cas<<":";        if(abs(sumb-sumw)<=1) {cout<<ans<<endl;continue;}        int now=0;        while(now<len){            int sum=0,pos=now;            for(int i=now;i<len;i++){                sum+=c[i];                if(sum==1||sum==-1||!sum) pos=i;            }            ans++;            now=pos+1;        }        cout<<ans-1<<endl;    }}

#include <cstdio>#include <iostream>#include <fstream>#include <string>#include <cstring>using namespace std;const int N = 1e5 + 10;string str;int GetAns(){    int n = str.length();    int sum = 0;    for (int i = 0; i < n; i++){        if (str[i] == '0') sum++;        else               sum--;    }    if (sum < 0) sum = -sum;    if (sum > 0) sum--;    return sum;}int main(){    int ca = 0;    while (cin >> str){        ca++;        int ans = GetAns();        cout << "Case #" << ca << ":" << ans << endl;    }    return 0;}