hdu6103-Kirinriki (2017杭电多校联赛第6场)

Kirinriki

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 796    Accepted Submission(s): 306


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

1
5
abcdefedcb

 

Sample Output

5

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

 

Source

2017 Multi-University Training Contest - Team 6

题目大意：给你一个计算两个相同长度的计算公式，每两个相同长度的字符串会得到一个值，现在给你一个字符串，然后给你一个值，询问在这个字符串中得到这个值的两个字符串的最大距离差。

解题思路：我们枚举每个位置，然后往两边寻找，暴力枚举每个位置得到最大值。

ac代码如下：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
const int N=50005;
int n,m,ans;
int q
,p
;
char s
;
void cal(int l,int r){
int x,y,v;
x=y=v=0;
q[y]=0;
p[y]=r-1;
y++;
while(l>=0&&r<n){
v+=abs(s[r]-s[l]);
q[y]=v;
p[y]=r;
y++;
while(v-q[x]>m)x++;
ans=max(ans,r-p[x]);
l--,r++;
}
}
int main(){
int T;
int i;
scanf("%d",&T);
while(T--){
scanf("%d",&m);
scanf("%s",s);
n=strlen(s);
ans=0;
for(i=1;i<n;i++){
cal(i-1,i);
cal(i-2,i);
}
printf("%d\n",ans);
}

return 0;
}

题目链接：点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=6103