Codeforces Round #370 (Div. 2) D. Memory and Scores 动态规划
2016-09-13 12:01
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D. Memory and Scores
题目连接:
http://codeforces.com/contest/712/problem/DDescription
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.Sample Input
1 2 2 1Sample Output
6Hint
题意
有两个人在玩游戏,一开始分数分别为a和b,每一局,每个人可以获得分数[-k,k]之间,问你A胜过B的方案数有多少种题解:
dp[i][j]表示第i轮之后,获得j分数的方案数。显然这个只会和上一轮有关,所以可以滚动数组优化,又显然可以前缀和优化。
然后维护一下DP
最后再枚举A的分数,统计一下答案就好了。
代码
#include<bits/stdc++.h> using namespace std; const int maxn = 4e5+7; const int le = 2e5; const int mod = 1e9+7; long long a,b,k,t; long long dp[2][maxn]; long long sum[maxn]; int now=0,pre=1; int main() { scanf("%lld%lld%lld%lld",&a,&b,&k,&t); dp[now][le]=1; for(int i=1;i<=t;i++) { for(int j=1;j<maxn;j++) { sum[j]=dp[now][j]+sum[j-1]; sum[j]%=mod; } swap(now,pre); memset(dp[now],0,sizeof(dp[now])); for(int j=1;j<maxn;j++) { dp[now][j]+=sum[min(maxn-1LL,j+k)]-sum[max(0LL,j-k-1)]; dp[now][j]%=mod; } } for(int j=1;j<maxn;j++) { sum[j]=dp[now][j]+sum[j-1]; sum[j]%=mod; } long long ans = 0; for(int j=0;j<maxn;j++) { ans += sum[a+j-b-1]%mod*dp[now][j]%mod; ans%=mod; } cout<<(ans+mod)%mod<<endl; }
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