gems gems gems

Posted 2020-10-08 不积跬步无以至千里，不积小流无以成江海

Now there are nn gems, each of which has its own value. Alice and Bob play a game with these nn gems.
They place the gems in a row and decide to take turns to take gems from left to right.
Alice goes first and takes 1 or 2 gems from the left. After that, on each turn a player can take kk or k+1k+1 gems if the other player takes kk gems in the previous turn. The game ends when there are no gems left or the current player can‘t take kk or k+1k+1 gems.
Your task is to determine the difference between the total value of gems Alice took and Bob took. Assume both players play optimally. Alice wants to maximize the difference while Bob wants to minimize it.

InputThe first line contains an integer TT (1≤T≤101≤T≤10 ), the number of the test cases.
For each test case:
the first line contains a numbers nn (1≤n≤200001≤n≤20000 );
the second line contains n numbers: V1,V2…VnV1,V2…Vn . (?100000≤Vi≤100000?100000≤Vi≤100000 )
OutputFor each test case, print a single number in a line: the difference between the total value of gems Alice took and the total value of gems Bob took.
Sample Input

1
3
1 3 2

Sample Output

4

#include<bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const int N = 20000 + 10;
const int K = 210;
const int TN = 255 + 10;
const int MOD = 255;
int T, n;
int v[N], pre[N], dp[2][TN][K];
int main()
{
    scanf("%d", &T);
    while(T-- && scanf("%d", &n)!=EOF)
    {
        memset(dp, 0, sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            scanf("%d", &v[i]);
            pre[i] = pre[i-1] + v[i];
        }
        int lmt = sqrt(n*2.0) + 1;
        for(int idx=n;idx;idx--)
        for(int k=lmt;k;k--)
        {
            if(idx+k <= n) {
                dp[0][idx&MOD][k] = pre[idx+k-1] - pre[idx-1] +
                    max(dp[1][(idx+k)&MOD][k], dp[1][(idx+k+1)&MOD][k+1] + v[idx+k]);
                dp[1][idx&MOD][k] = -pre[idx+k-1] + pre[idx-1] + 
                    min(dp[0][(idx+k)&MOD][k], dp[0][(idx+k+1)&MOD][k+1] - v[idx+k]);
            }   
            else if(idx+k-1 <= n) {
                dp[0][idx&MOD][k] = dp[1][(idx+k)&MOD][k] + pre[idx+k-1] - pre[idx-1];
                dp[1][idx&MOD][k] = dp[0][(idx+k)&MOD][k] - pre[idx+k-1] + pre[idx-1];
            }
        }
        printf("%d\n", dp[0][1][1]);
    }
}