2017 Multi-University Training Contest

Problem Description

We define the distance of two strings A and B with same length n is
disA,B=∑i=0n−1|Ai−Bn−1−i|
The difference between the two characters is defined as the difference in ASCII.
You should find the maximum length of two non-overlapping substrings in given string S, and the distance between them are less then or equal to m.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integers m : the limit distance of substring.
Then a string S follow.

Limits
T≤100
0≤m≤5000
Each character in the string is lowercase letter, 2≤|S|≤5000
∑|S|≤20000

 

Output

For each test case output one interge denotes the answer : the maximum length of the substring.

 

Sample Input

15abcdefedcb

 

Sample Output

5
Hint
[0, 4] abcde[5, 9] fedcbThe distance between them is abs('a' - 'b') + abs('b' - 'c') + abs('c' - 'd') + abs('d' - 'e') + abs('e' - 'f') = 5

#include <iostream>#include <stdio.h>#include <string.h>#include <queue>#include <cmath>#include <stdlib.h>using namespace std;typedef long long LL;const LL MAX = 5000+10;const LL mod = 7654321; const double eps = 1e-8;const double PI = acos(-1.0);#define zero(x) (((x)>0?(x):(-x))<eps)int ans, m, len;char s[MAX];void solve(int a, int b){    int l = 0, r = 0, sum = 0;    while(a+r<b-r){        sum+=abs(s[a+r]-s[b-r]);        r++;        if(b-l-a-l+1<=ans*2)break;        //printf("%d %d\n", l, r);        //printf("%d %d %d %d\n", a+l, a+r, b-r, b-l);        while(sum>m&&r>l){            sum-=abs(s[a+l]-s[b-l]);            l++;        }        if(ans<r-l)ans = r-l;    }}int main() {    int t;    scanf("%d", &t);    while(t--){        scanf("%d", &m);        scanf("%s", s);        len = strlen(s);        ans = 0;        for(int i = 1; i<len; ++i){            solve(0, i);        }        for(int i = 0; i<len-1; ++i){            solve(i,len-1);        }        printf("%d\n", ans);    }    return 0;}