Traveling on the Axis 模拟

题目描述
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 ti (ti ∈{ 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

where both p and q are integers. Can you help him?

输入
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|≤105, |s| = n, si ∈{‘0’, ‘1’} for all 1≤i≤|s| ),indicating the types of the traffic lights. If si = ‘0’, the traffic light at point (i-0.5) is of type 0 and is initially red; If si = ‘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 106.

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

样例输入
3
101
011
11010
样例输出
12
15
43

提示
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.

一道比较复杂的模拟，按照题目先手写一个测试用例的全过程

1 1 0 1 0
0 0 1 0 1
0出发: 1 3 4 5 6  19  11
1出发：  1 2 3 4  10  10
2出发：    2 3 4  9   01
3出发：      1 2  3   10
4出发：        2  2   0

首先，根据相邻两个点的情况，不难发现，如果两个相邻的红绿灯的初始状态相同，那么通过一个红绿灯之后，下一个一定会拦下，基于这个，我们先算出从0出发的所有情况。之后从不同位置出发的所有情况，不难发现每往后一个位置，都会减去最前面的数据，这是因为出发位置的改变使情况减少了，更换起点之后，应该根据起点位置，再做新的初试调整，也就是说如果修改位置后前面两个为11、10、01、00，都应该对应不同的操作，以11为例，如果调整前最前面两个为11，那么初始位置向后一一个后变成10，就减少了开始等待的两分钟，直接通过两个路口，这引起整个后续情况都少了两分钟，其余情况以此类推。

另外几个数据需要定义成long long int，不然会WA。

AC代码

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{int t;scanf("%d",&t);getchar();while(t--){string str;cin>>str;getchar();long long int time=0,ans=0;if(str[0]=='0')time=2;else if(str[0]=='1')time=1;ans+=time;for(int i=1;i<str.length();i++){if(str[i]==str[i-1])time+=2;elsetime++;ans+=time;    }long long int sum=ans;int m=str.length();for(int i=1;i<str.length();i++){if(str[i-1]=='1'&&str[i]=='1')//11ans=ans-((m-1)*2+1);else if(str[i-1]=='1'&&str[i]=='0')//10ans=ans-1;else if(str[i-1]=='0')//0Xans=ans-2*m;sum+=ans;m--;}printf("%lld\n",sum);}return 0;
}

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