zoukankan      html  css  js  c++  java
  • Codeforces Round #302 (Div. 2) C 简单dp

    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 nmbmod (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
     1 #pragma comment(linker, "/STACK:1024000000,1024000000")
     2 #include<iostream>
     3 #include<cstdio>
     4 #include<cmath>
     5 #include<string>
     6 #include<queue>
     7 #include<algorithm>
     8 #include<stack>
     9 #include<cstring>
    10 #include<vector>
    11 #include<list>
    12 #include<set>
    13 #include<map>
    14 #include<bitset>
    15 #include<time.h>
    16 using namespace std;
    17 int dp[505][505];
    18 int n,m,b,mod;
    19 int main(){
    20     scanf("%d %d %d %d",&n,&m,&b,&mod);
    21     dp[0][0]=1;
    22     int exm;
    23     for(int i=1;i<=n;i++){
    24         scanf("%d",&exm);
    25         for(int j=1;j<=m;j++){
    26             for(int k=exm;k<=b;k++)
    27                 dp[j][k]=(dp[j][k]+dp[j-1][k-exm])%mod;
    28         }
    29     }
    30     int ans=0;
    31     for(int i=0;i<=b;i++)
    32         ans=(ans+dp[m][i])%mod;
    33     printf("%d
    ",ans);
    34     return 0;
    35 }
  • 相关阅读:
    UILabel的使用
    CGAffineTransform的使用
    UIView的常用方法
    UICollectionViewController的用法1
    网址连接
    android developers blog
    Java并发编程:volatile关键字解析
    Android触摸屏事件派发机制详解与源码分析
    setScale,preScale 和 postScale 的区别
    android 内存
  • 原文地址:https://www.cnblogs.com/hsd-/p/7309767.html
Copyright © 2011-2022 走看看