Codeforces Round #370 (Div. 2) D. Memory and Scores DP

Posted 2020-08-09 Meek

D. Memory and Scores

 

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

input

 

1 2 2 1

output

 

6

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 are3 + 2 + 1 = 6 possible games in which Memory wins.

 

 题意：

　　A,B两人玩t轮游戏

　　每轮游戏没人可以从[-k,k]中获取任意的一个分数

　　AB起始分数分别为a,b

　　问你最终A分数严格比B多的方案数

题解：

　　设定dp[i][j]为第i轮 获得分数j的方案数

　　这个可以进行滚动数组和前缀和优化

　　最后枚举一个人的 分数 得到答案

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<set>
using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair

typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 5e5+10, M = 1e2+11, mod = 1e9+7, inf = 2e9;

int a,b,k,t;
LL sum[N], dp[2][N];
int main() {
        scanf("%d%d%d%d",&a,&b,&k,&t);
        int now = 1;
        int last = now ^ 1;
        dp[last][0] = 1;
        for(int i = 0; i <= 2 * k * t; ++i) sum[i] = 1;
        for(int i = 1; i <= t; ++i) {
            for(int  j = 0; j <= 2*k*t; ++j) {
                if(j <= 2 * k) dp[now][j] = sum[j];
                else {
                    dp[now][j] = (( sum[j] - sum[j - 2*k - 1] ) % mod + mod ) % mod;
                }
            }
            sum[0] = dp[now][0];
            for(int j = 1; j <= 4 * k * t; ++j)
                sum[j] = ((sum[j-1] + dp[now][j]) % mod + mod) % mod;
            now^=1;
        }
        LL ans = 0;
        for(int i = 0; i <= 2 * k * t; ++i) {
            if(a + i - 1 - b >= 0)ans = (ans + dp[now^1][i] * sum[a + i - 1 - b]%mod) % mod;
        }
        cout<<(ans+mod) % mod<<endl;
        return 0;
}