Codeforces 712D Memory and Scores【dp+前缀和】

D. Memory and Scores time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard 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 Input 1 2 2 1 Output 6 Input 1 1 1 2 Output 31 Input 2 12 3 1 Output 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.

题目大意：

A一开始的分数是a，B一开始的分数是b，问最终A比B分数高的方案数。

一共t轮游戏，每轮游戏每个人可以在【-k，k】之间选一个数和自己的分数进行累加。

思路：

设定dp【i】【j】表示玩了i轮，分数为j的方案数，那么明显有：dp【i】【j】=dp【i-1】【j+v】（-k<=v<=k）那么这种题就很无趣了，直接维护一个前缀和的套路展现无疑啊。

那么维护两个选手的方案数，对应结果：

output+=ans【j】*ans2【j-v】（v>=1）；

显然直接枚举还是会TLE ，那么继续维护一个前缀和即可。

Ac代码：

#include<stdio.h> #include<string.h> using namespace std; #define mod 1000000007 #define ll __int64 ll dp[105][300500]; ll sum[105][300500]; ll ans[300500]; ll ans2[300500]; ll preans2[300500]; ll kuaisucheng(ll a,ll b) { ll ans=0; while(b) { if(b&1) { ans=(ans+a); if(ans>=mod)ans-=mod; } a=(a+a); if(a>=mod)a-=mod; b/=2; } return ans; } int main() { ll a,b,k,t; while(~scanf("%I64d%I64d%I64d%I64d",&a,&b,&k,&t)) { int mid=150000; memset(sum,0,sizeof(sum)); memset(dp,0,sizeof(dp)); dp[0][a+mid]=1; for(int j=4000;j<=260000;j++)sum[0][j]=sum[0][j-1]+dp[0][j],sum[0][j]%=mod; for(int i=1;i<=t;i++) { for(int j=4000;j<=260000;j++) { if(j+k<=260000) dp[i][j]=(sum[i-1][j+k]-sum[i-1][j-k-1]+mod)%mod; } for(int j=4000;j<=260000;j++)sum[i][j]=sum[i][j-1]+dp[i][j],sum[i][j]%=mod; } for(int j=4000;j<=260000;j++)ans[j]=dp[t][j]; memset(sum,0,sizeof(sum)); memset(dp,0,sizeof(dp)); dp[0][b+mid]=1; for(int j=4000;j<=260000;j++)sum[0][j]=sum[0][j-1]+dp[0][j],sum[0][j]%=mod; for(int i=1;i<=t;i++) { for(int j=4000;j<=260000;j++) { if(j+k<=260000) dp[i][j]=(sum[i-1][j+k]-sum[i-1][j-k-1]+mod)%mod; } for(int j=4000;j<=260000;j++)sum[i][j]=sum[i][j-1]+dp[i][j],sum[i][j]%=mod; } for(int j=4000;j<=260000;j++)ans2[j]=dp[t][j]; ll output=0; for(int j=4500;j<=255000;j++)preans2[j]=preans2[j-1]+ans2[j],preans2[j]%=mod; for(int j=4500;j<=255000;j++)output+=kuaisucheng(preans2[j-1],ans[j]),output%=mod; printf("%I64d\n",output); } }