Traveling on the Axis
Time Limit: 1 Second Memory Limit: 65536 KB
BaoBao is taking a walk in the interval on the number axis, but he is not free to move, as at every point for all , where is an integer, stands a traffic light of type ().
BaoBao decides to begin his walk from point and end his walk at point (both and are integers, and ). During each unit of time, the following events will happen in order:
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Let's say BaoBao is currently at point , he will then check the traffic light at point . If the traffic light is green, BaoBao will move to point ; If the traffic light is red, BaoBao will remain at point .
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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 as the total units of time BaoBao needs to move from point to point . For some reason, BaoBao wants you to help him calculate
where both and are integers. Can you help him?
Input
There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:
The first and only line contains a string (, , for all ), indicating the types of the traffic lights. If , the traffic light at point is of type 0 and is initially red; If , the traffic light at point is of type 1 and is initially green.
It's guaranteed that the sum of of all test cases will not exceed .
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 , , , , and , so the answer is .
For the second sample test case, it's easy to calculate that , , , , and , so the answer is .
Author: WENG, Caizhi
Source: The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online
题意：走红绿灯，红绿灯每秒变一次，问走到最后要多少秒。求的是从任一点走到最后的和。
解题思路：首先先预处理从开始走到每一个点所需要的时间，然后求一个后缀和。对于任一点，它第一个红绿灯所需要的时间要么是1s要么是2s，所以根据这个判断，用后缀和减去多出来的时间就好了。
#include <iostream>
#include <string.h>
#include <math.h>
#include <vector>
#include <algorithm>
#include <stdio.h>
#include <queue>
using namespace std;
const int MAXN = 100010;
const int INF = 0x3f3f3f3f;
typedef long long ll;
int a[MAXN];
ll sum[MAXN];
char str[MAXN];
int main()
{
int TT;
scanf("%d",&TT);
while (TT--)
{
memset(a,0,sizeof(a));
memset(sum,0,sizeof(sum));
scanf("%s",str);
int len=strlen(str);
ll cur=0;
bool cha=0;
for(int i=0;i<len;i++){
if(str[i]=='1'&&cha==0){
cur++;
a[i+1]=cur;
cha=1;
continue;
}
if(str[i]=='1'&&cha==1){
cur++;
cur++;
a[i+1]=cur;
continue;
}
if(str[i]=='0'&&cha==0){
cur++;
cur++;
a[i+1]=cur;
continue;
}
if(str[i]=='0'&&cha==1){
cur++;
a[i+1]=cur;
cha=0;
continue;
}
}
for(int i=len;i>=1;i--)
sum[i]=sum[i+1]+a[i];
ll ans=0;
for(int i=1;i<=len;i++){
if(str[i-1]=='0'){
ll sub=a[i]-2;
ans+=sum[i]-sub*(len-i+1);
}
else{
ll sub=a[i]-1;
ans+=sum[i]-sub*(len-i+1);
}
}
printf("%lld\n",ans);
}
return 0;
}