数据结构作业3-4（周）问题F：Turn off the light（关下灯）

题目描述

There are n lights aligned in a row. These lights are numbered 1 to n from left to right.
Initially some of the lights are turned on. Chiaki would like to turn off all the lights.
Chiaki starts from the p-th light. Each time she can go left or right (i.e. if Chiaki is at x, then she can go to x−1 or x+1) and then press the switch of the light in that position (i.e. if the light is turned on before, it will be turned off and vise versa).
For each p=1,2,…,n, Chiaki would like to know the minimum steps needed to turn off all the lights.

///一排中有n个灯，从左到右依次标号1到n，起初有些灯是亮着的。千明小姐想要关掉所有的灯。千明小姐从第p盏灯开始，每次她可以左右横跳一格来改变那个格子中灯的状态（关上的就打开，开着的就关上）。对于每个p=1,2，…，n，千明小姐想知道关闭所有灯所需的最小步骤。///

输入

There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:
The first line contains an integer n (2≤n≤106) – the number of lights.
The second line contains a binary string s where si=1 means the i-th light is turned on and si=0 means i-th light is turned off.
It is guaranteed that the sum of all n does not exceed 107.

///有多个测试用例。输入的第一行是一个整数T，表示测试用例的数量。
对于每个测试用例：第一行包含整数n（2≤n≤106）——灯光数。
第二行包含一个二进制字符串s，其中si=1表示第i个灯打开，si=0表示第i个灯关闭。题中会保证n<107。///

输出

最终的结果需要计算i*zi (i为1~字符串的长度的范围，zi为p在i的最小步骤），从1到n的和
输出和%(1e9+7)

举例：对于测试用例的111来说，p=1，则需要经过111>101>001>011>010>000 5个步骤；p=2，需经过111>011>001>000 3个步骤；p=3，需经过跟p=1一样的5个步骤，只不过倒过来。则结果为：15+23+3*5=26

样例输入

样例输出

0
26
432

你说这题该怎么办？我是真的不会。
你们自己去看大佬写的吧：运用了异或运算等知识

https://blog.csdn.net/BinGoo0o0o/article/details/81190093

https://blog.csdn.net/qq_18869763/article/details/81193622

最终代码：

#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
#define N 1010000
const int mod = 1e9+7;
int n,T;
bool f[N],g[N],preg[N],preg2[N];
char s[N];int ret[N],ret2[N],cnt[N],cnt2[N];
void work(bool *g,int *ret,int n)
{   for(int i = 1; i <= n; i++)         {ret[i]=1<<30;}int fo=n+1,lo=-1;    int i=1;    while(i<=n)    {   if(g[i])         {if (fo==n+1) fo=i;lo=i;}        i++;    }    if (fo>lo)     {   i=1;        while(i<=n)        {   ret[i]=0;           i++;        }         return;    }    for(i = 1; i <= n; i++)     {      preg[i]=preg[i-1]^g[i]^1;        preg2[i]=preg2[i-1]^g[i];        cnt[i]=cnt[i-1]+preg[i];        cnt2[i]=cnt2[i-1]+preg2[i];    }    for( i = 1; i <= lo; i++)     {      if (i<=fo)        {   int fw=i,lw=lo;            if (i==lo)             {   ret[i]=min(ret[i],3);                continue;            }            int ans=lw-fw;            if (preg[fw-1])ans+=2*(cnt[lw-1]-cnt[fw-1]);            else ans+=2*(lw-fw-cnt[lw-1]+cnt[fw-1]);            if (preg[lw-1]^preg[fw-1]^1) ans--;            ret[i]=min(ret[i],ans);        }         else        {      int fw=fo,lw=lo;            int ans=i-fo+lw-fw;            if (preg2[fw-1])                 ans += 2*(cnt2[i-1]-cnt2[fw-1]);            else    ans+=2*(i-fw-cnt2[i-1]+cnt2[fw-1]);            int x=preg2[i-1]^preg2[fw-1]^1;            if (x==preg[i-1])                 ans+=2*(cnt[lw-1]-cnt[i-1]);            else  ans+=2*(lw-i-cnt[lw-1]+cnt[i-1]);            if (x^preg[i-1]^preg[lw-1])   ans--;            ret[i]=min(ret[i],ans);        }    }
}
int main()
{       scanf("%d",&T);    while(T--)    {    scanf("%d",&n);        scanf("%s",s+1);        int i=1;        while(i<=n)        {       g[i]=(s[i]=='1');            i++;        }        work(g,ret,n);        reverse(g+1,g+n+1);        work(g,ret2,n);        LL ans=0;        for(i = 1; i <= n; i++)         {      ret[i]=min(ret[i],ret2[n+1-i]);            ans=(ans+(LL)i*ret[i])%mod;        }        printf("%lld\n",ans);    }    return 0;
}