D. Memory and Scores [dp]_a. matching time limit per test2 seconds memory li

D. Memory and Scores
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
Memory and his friend Lexa are competing to get higher score in one popular computer game. Memory starts with score a and Lexa starts with score b. In a single turn, both Memory and Lexa get some integer in the range [ - k;k] (i.e. one integer among  - k,  - k + 1,  - k + 2, …,  - 2,  - 1, 0, 1, 2, …, k - 1, k) and add them to their current scores. The game has exactly t turns. Memory and Lexa, however, are not good at this game, so they both always get a random integer at their turn.

Memory wonders how many possible games exist such that he ends with a strictly higher score than Lexa. Two games are considered to be different if in at least one turn at least one player gets different score. There are (2k + 1)2t games in total. Since the answer can be very large, you should print it modulo 109 + 7. Please solve this problem for Memory.

Input
The first and only line of input contains the four integers a, b, k, and t (1 ≤ a, b ≤ 100, 1 ≤ k ≤ 1000, 1 ≤ t ≤ 100) — the amount Memory and Lexa start with, the number k, and the number of turns respectively.

Output
Print the number of possible games satisfying the conditions modulo 1 000 000 007 (109 + 7) in one line.

Examples
inputCopy
1 2 2 1
outputCopy
6
inputCopy
1 1 1 2
outputCopy
31
inputCopy
2 12 3 1
outputCopy
0
Note
In the first sample test, Memory starts with 1 and Lexa starts with 2. If Lexa picks  - 2, Memory can pick 0, 1, or 2 to win. If Lexa picks  - 1, Memory can pick 1 or 2 to win. If Lexa picks 0, Memory can pick 2 to win. If Lexa picks 1 or 2, Memory cannot win. Thus, there are 3 + 2 + 1 = 6 possible games in which Memory wins.

参考了dalao的思路链接
写了前缀和的代码

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 210010;
const int mod = 1e9+7;
int dp[110][maxn],sum[maxn];//dp[i][j]代表第i轮得分为j的方案数
int a,b,k,t;
 
int main()
{
    scanf("%d%d%d%d",&a,&b,&k,&t);
    dp[0][0] = 1;
    for(int i=1; i<=t; i++){
        for(int j=0; j<maxn; j++) 
            sum[j] = ((j>0?sum[j-1]:0) + dp[i-1][j])%mod;//dpi-1的前缀和
        for(int j=0; j<maxn; j++) 
            dp[i][j] = (sum[j]-(j-2*k-1>=0?sum[j-2*k-1]:0) + mod)%mod;//在整个数组范围内更新dpi
    }
    for(int i=0; i<maxn; i++)
        sum[i] = ((i>0?sum[i-1]:0) + dp[t][i])%mod;
    //枚举a的得分
    int ans = 0;
    for(int i=0; i<=2*k*t; i++){
        (ans += (1LL) * dp[t][i] * (i+a-b-1<0?0:sum[i+a-b-1]) % mod) %= mod;
    }
    printf("%d\n",ans);
    return 0;
}
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