牛客周赛41 D(小红的好串)

题目大意:
给你一个字符串，有q次询问，每次询问给一个区间l到r，问最少修改多少次使得区间内的字符串是好串
好串的定义:长度和自身相同的,拥有red子序列最多的字符串

题目分析:
可以发现red子序列最多的情况一定是rrr…ee…ddd…这样子
可以分三种情况:设字符串长度为len
if($len$%3==0):
r,e,d的个数均分，用前缀和查询和理想情况差多少即可

if($len$%3==1):
有三种情况，r多一个,e多一个，d多一个,求个min

if($len$%3==2)
也有三种情况(r,e)多一个，（r,d）多一个,(e,d)多一个,也求个min

#include<bits/stdc++.h>
using namespace std;
using i64 = long long;
#define ios ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);

//len小于3就是好串
int n,q;
string s;
i64 r[200010],e[200010],d[200010];

int main(){
    ios;
    cin>>n>>q>>s;
    s = ' '+s;
    for(int i = 1;i<=n;++i){
        r[i] = r[i-1]+(s[i]=='r');
        e[i] = e[i-1]+(s[i]=='e');
        d[i] = d[i-1]+(s[i]=='d');
    }
    while(q--){
        int l,ri;
        cin>>l>>ri;
        if(ri-l+1<3) cout<<"0\n";
        else if((ri-l+1)%3==0){
            int k = (ri-l+1)/3;
            cout<<k*3LL-(r[l+k-1]-r[l-1])-(e[l+k+k-1]-e[l+k-1])-(d[ri]-d[l+k+k-1])<<"\n";
        }else if((ri-l+1)%3==1){
            i64 mi = 0x3f3f3f3f;
            int k = (ri-l+1)/3;
            mi = min(mi,k+k+(k+1)-(r[l+k]-r[l-1])-(e[l+k+k]-e[l+k])-(d[ri]-d[l+k+k]));
            cout<<r[l+k]-r[l-1]<<" "<<e[l+k+k]-e[l+k]<<" "<<d[ri]-d[l+k+k]<<"\n";
            mi = min(mi,k+k+(k+1)-(r[l+k-1]-r[l-1])-(e[l+k+k]-e[l+k-1])-(d[ri]-d[l+k+k]));
            mi = min(mi,k+k+(k+1)-(r[l+k-1]-r[l-1])-(e[l+k+k-1]-e[l+k-1])-(d[ri]-d[l+k+k-1]));
            cout<<mi<<"\n";
        }else if((ri-l+1)%3==2){
            i64 mi = 0x3f3f3f3f;
            int k = (ri-l+1)/3;
            mi = min(mi,k+(k+1)+(k+1)-(r[l+k]-r[l-1])-(e[l+k+k+1]-e[l+k])-(d[ri]-d[l+k+k+1]));
            mi = min(mi,k+(k+1)+(k+1)-(r[l+k]-r[l-1])-(e[l+k+k]-e[l+k])-(d[ri]-d[l+k+k]));
            mi = min(mi,k+(k+1)+(k+1)-(r[l+k-1]-r[l-1])-(e[l+k+k]-e[l+k-1])-(d[ri]-d[l+k+k]));
            cout<<mi<<"\n";
        }
    }
    return 0;
}

总的来说如果想到了就是一道靠代码基本功的题目