The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online H Traveling on the Axis ——dp

H Traveling on the Axis
BaoBao is taking a walk in the interval [0,n] on the number axis, but he is not free to move, as at every point (i−0.5) for all i∈[1,n], where i is an integer, stands a traffic light of type t
​i (t​i∈{0,1}).

BaoBao decides to begin his walk from point p and end his walk at point q (both p and q are integers, and $p<q$). During each unit of time, the following events will happen in order:

Let’s say BaoBao is currently at point x, he will then check the traffic light at point (x+0.5). If the traffic light is green, BaoBao will move to point (x+1); If the traffic light is red, BaoBao will remain at point x.
All the traffic lights change their colors. If a traffic light is currently red, it will change to green; If a traffic light is currently green, it will change to red.
A traffic light of type 0 is initially red, and a traffic light of type 1 is initially green.

Denote t(p,q) as the total units of time BaoBao needs to move from point p to point q. For some reason, BaoBao wants you to help him calculate

​p=0
​∑
​n−1
​​
​q=p+1
​∑
​n
​​ t(p,q)
where both p and q are integers. Can you help him?

Input
There are multiple test cases. The first line of the input contains an integer T, indicating the number of test cases. For each test case:

The first and only line contains a string s (1≤∣s∣≤10
​5
​​ , ∣s∣=n, s
​i
​​ ∈{‘0’,‘1’} for all 1≤i≤∣s∣), indicating the types of the traffic lights. If s
​i
​​ =‘0’, the traffic light at point (i−0.5) is of type 0 and is initially red; If s
​i
​​ =‘1’, the traffic light at point (i−0.5) is of type 1 and is initially green.

It’s guaranteed that the sum of ∣s∣ of all test cases will not exceed 10
​6
​​ .

Output
For each test case output one line containing one integer, indicating the answer.

Sample Input
3
101
011
11010
Sample Output
12
15
43
Hint
For the first sample test case, it’s easy to calculate that t(0,1)=1, t(0,2)=2, t(0,3)=3, t(1,2)=2, t(1,3)=3 and t(2,3)=1, so the answer is 1+2+3+2+3+1=12.

For the second sample test case, it’s easy to calculate that t(0,1)=2, t(0,2)=3, t(0,3)=5, t(1,2)=1, t(1,3)=3 and t(2,3)=1, so the answer is 2+3+5+1+3+1=15.
题目链接：https://pintia.cn/problem-sets/1036903825309761536/problems/1041156323504345088
题意：从0走到n，0.5，1.5,2.5。。。的位置有红绿灯，给你一个串，代表初始的每个位置是红灯还是绿灯，0代表红灯，1代表绿灯，问你从0走到1，从0走到2。。。从0走到n，从1走到2。。。从1走到n。。。的累加
题解：
我是用了dp，从后往前推，因为后面的状态是可以记录下来的而且每个点只有两个。
dp[0][i]代表第i个状态的当前状态，dp[1][i]代表当前状态的反状态。
那么我们可以知道如果当前点为0，那么我们需要等2秒才到下一个状态，而下一个状态就是当前状态的下一个，就是下一个状态加上2乘数的个数。如果是1的话，就是下一个状态的反加上1乘有多少个数举个例子：
比如说01:一开始是1，那么dp[0][2]=1,dp[1][2]=2
接下来是0，dp[0][1]=5,dp[1][1]=3
最后全部加起来就好了。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
char a[100005];
ll dp[2][100005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",a+1);
        int n=strlen(a+1);
        if(a[n]=='1')   dp[1][n]=2,dp[0][n]=1;
        else dp[1][n]=1,dp[0][n]=2;
        for(int i=n-1;i>=1;i--)
        {
            if(a[i]=='1')
            {
                dp[0][i]=dp[1][i+1]+n-i+1;
                dp[1][i]=dp[1][i+1]+2*(n-i+1);
            }
            else
            {
                dp[0][i]=dp[0][i+1]+2*(n-i+1);
                dp[1][i]=dp[0][i+1]+n-i+1;
            }
        }
        ll ans=0;
        for(int i=1;i<=n;i++)
            ans+=dp[0][i];//printf("dp[0][%d] = %lld ,dp[1][%d] = %lld\n",i,dp[0][i],i,dp[1][i]),
        printf("%lld\n",ans);
    }
    return 0;
}