Codeforces 1315B Homecoming （二分）

After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are

The crossroads are represented as a string

If for two crossroads

For example, if

$\text{[math]}$

buy one bus ticket to get from
buy one tram ticket to get from
buy one bus ticket to get from

Thus, in total he needs to spend

Now Petya is at the first crossroad, and he wants to get to the

Help him to choose the closest crossroad

Input

Each test contains one or more test cases. The first line contains the number of test cases

The first line of each test case consists of three integers

The second line of each test case consists of one string

It is guaranteed, that the sum of the length of strings

Output

For each test case print one number — the minimal index

Example

Input

5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB

Output

2
1
3
1
6
大意就是这个人有p元，公交车在乘车区间a元，有轨电车b元，要到达n的话钱不一定够，问先步行再坐交通工具的话最少需要走几站。从后往前考虑比较方便。根据题意，其实s[n]是什么不用管，因为要到达的就是s[n]，所以分别求出s[n]到s[i](0<=i<=n-1,这里是字符串下标）所需要花的钱。容易看出花费构成的序列是单调的因此直接从里面找第一个小于等于p的数对应的位置输出即可，没有的话直接输出最后一个位置。理论上这里可以直接二分lower_bound,不知道为什么这么写会挂在第56个点...最后O(n)竟然也能过...还请大佬指教。

#include <bits/stdc++.h>
using namespace std;
long long a,b,p;
char s[1000005];
unsigned long long n[1000005];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        scanf("%lld%lld%lld",&a,&b,&p);
        long long i,pre=0;
        scanf("%s",s);
        for(i=strlen(s)-2;i>=0;i--)//不用管最后一个是啥 
        {
            long long temp=0;
            if(s[i]=='A')temp=a;
            else if(s[i]=='B')temp=b;
            if(i==strlen(s)-2)
            {
                pre=temp;
                n[i]=temp;
                continue;
            }
            if(s[i]==s[i+1])n[i]=n[i+1];
            else
            {
                pre=n[i+1];
                n[i]=pre+temp;
            }
        }
//        long long pos=lower_bound(n,n+strlen(s)-1,p,greater<int>())-n;//找到递减序列里第一个小于等于p的数 //为啥二分过不了？？？？？？？ 
//        if(pos==strlen(s))
//        {
//            cout<<strlen(s)<<endl;
//            continue;
//        }
//        cout<<pos+1<<endl;    
        int pos=0;    
        bool flag=0;
        for(i=strlen(s)-2;i>=0;i--)
        {
            if(i==0)
            {
                if(n[0]<=p)
                {
                    pos=1;
                    flag=1;
                }
            }
            if(n[i]<=p&&n[i-1]>p)
            {
                pos=i+1;
                flag=1;
                break;
            }
        }
        if(!flag)pos=strlen(s);
        cout<<pos<<endl;
    }
    return 0;
 }