Codeforces Round #302 (Div. 2) C 简单dp
2017-08-09 00:03
375 查看
C. Writing Code
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
input
output
input
output
input
output
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
input
3 3 3 100 1 1 1
output
10
input
3 6 5 1000000007 1 2 3
output
0
input
3 5 6 11 1 2 1
output
0
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> #include<bitset> #include<time.h> using namespace std; int dp[505][505]; int n,m,b,mod; int main(){ scanf("%d %d %d %d",&n,&m,&b,&mod); dp[0][0]=1; int exm; for(int i=1;i<=n;i++){ scanf("%d",&exm); for(int j=1;j<=m;j++){ for(int k=exm;k<=b;k++) dp[j][k]=(dp[j][k]+dp[j-1][k-exm])%mod; } } int ans=0; for(int i=0;i<=b;i++) ans=(ans+dp[m][i])%mod; printf("%d\n",ans); return 0; }
相关文章推荐
- codeforces round# 302 (div1 C) (状压dp)
- Codeforces Round #260 (Div. 1) A. Boredom (简单dp)
- Codeforces Round #267 (Div. 2)D(DFS+单词hash+简单DP)
- Codeforces Round #343 (Div. 2) C. Famil Door and Brackets(简单dp)
- 【Codeforces Round 367 (Div 2) C】【简单DP】Hard problem
- Codeforces Round #302 (Div. 2) E. Remembering Strings(状压dp)
- 【Codeforces Round 363 (Div 2) C】【简单DP】Vacations 一天运动 一天学习最少休息日数
- Codeforces Round #272 (Div. 2) B Dreamoon and WiFi(简单DP)
- Codeforces Round #239 (Div. 2) A. Line to Cashier(简单题)
- Codeforces Round #106(Div. 2) 149D. Coloring Brackets 区间DP 记忆化搜索
- Codeforces Round #144 (Div. 1), problem: (B) Table DP 组合数学
- Codeforces Round #341 (Div. 2) E. Wet Shark and Blocks(矩阵优化DP)
- Codeforces Round #Pi (Div. 2) C. Geometric Progression dp
- Codeforces Round #339 (Div. 2)C. Peter and Snow Blower(简单几何)
- codeforces round 362 div2 D Puzzles 树形dp + 期望
- Codeforces Round #460 (Div. 2) D. Substring BFS、拓扑排序、dp
- Codeforces Round #382 (Div. 1) C. Ostap and Tree(树形DP)
- 【Codeforces Round 363 (Div 2) E】【概率DP 期望DP 逆推等价法】LRU Cache替换LRU原则超多步数后每个数据在Cache中的概率
- Codeforces Round #362 (Div. 2) D 树形dp
- Codeforces Round #144 (Div. 1) B dp