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.
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.
Print the number of possible games satisfying the conditions modulo 1 000 000 007 (109 + 7) in one line.
1 2 2 1
6
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+, M = 1e2+, mod = 1e9+, inf = 2e9; int a,b,k,t;
LL sum[N], dp[][N];
int main() {
scanf("%d%d%d%d",&a,&b,&k,&t);
int now = ;
int last = now ^ ;
dp[last][] = ;
for(int i = ; i <= * k * t; ++i) sum[i] = ;
for(int i = ; i <= t; ++i) {
for(int j = ; j <= *k*t; ++j) {
if(j <= * k) dp[now][j] = sum[j];
else {
dp[now][j] = (( sum[j] - sum[j - *k - ] ) % mod + mod ) % mod;
}
}
sum[] = dp[now][];
for(int j = ; j <= * k * t; ++j)
sum[j] = ((sum[j-] + dp[now][j]) % mod + mod) % mod;
now^=;
}
LL ans = ;
for(int i = ; i <= * k * t; ++i) {
if(a + i - - b >= )ans = (ans + dp[now^][i] * sum[a + i - - b]%mod) % mod;
}
cout<<(ans+mod) % mod<<endl;
return ;
}