zoukankan      html  css  js  c++  java
  • poj3046

    Description

    Bessie was poking around the ant hill one day watching the ants march to and fro while gathering food. She realized that many of the ants were siblings, indistinguishable from one another. She also realized the sometimes only one ant would go for food, sometimes a few, and sometimes all of them. This made for a large number of different sets of ants!

    Being a bit mathematical, Bessie started wondering. Bessie noted that the hive has T (1 <= T <= 1,000) families of ants which she labeled 1..T (A ants altogether). Each family had some number Ni (1 <= Ni <= 100) of ants.

    How many groups of sizes S, S+1, ..., B (1 <= S <= B <= A) can be formed?

    While observing one group, the set of three ant families was seen as {1, 1, 2, 2, 3}, though rarely in that order. The possible sets of marching ants were:

    3 sets with 1 ant: {1} {2} {3}
    5 sets with 2 ants: {1,1} {1,2} {1,3} {2,2} {2,3}
    5 sets with 3 ants: {1,1,2} {1,1,3} {1,2,2} {1,2,3} {2,2,3}
    3 sets with 4 ants: {1,2,2,3} {1,1,2,2} {1,1,2,3}
    1 set with 5 ants: {1,1,2,2,3}

    Your job is to count the number of possible sets of ants given the data above.

    Input

    * Line 1: 4 space-separated integers: T, A, S, and B

    * Lines 2..A+1: Each line contains a single integer that is an ant type present in the hive

    Output

    * Line 1: The number of sets of size S..B (inclusive) that can be created. A set like {1,2} is the same as the set {2,1} and should not be double-counted. Print only the LAST SIX DIGITS of this number, with no leading zeroes or spaces.

    Sample Input

    3 5 2 3
    1
    2
    2
    1
    3

    Sample Output

    10

    Hint

    INPUT DETAILS:

    Three types of ants (1..3); 5 ants altogether. How many sets of size 2 or size 3 can be made?


    OUTPUT DETAILS:

    5 sets of ants with two members; 5 more sets of ants with three members
     
    dp[i][j]表示前i类选0-j个方案数
    则转移方程为dp[i][j]=dp[i][j-1]-dp[i-1][j-num[i]-1]+dp[i-1][j]
    用滚动数组将空间压为一维
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    
    const int mod = 1e6;
    
    int n,m,s,b,num[1100],f[100005];
    
    int main(){
        while(scanf("%d%d%d%d",&n,&m,&s,&b)!=EOF){
            memset(num,0,sizeof(num));
            for(int i=1;i<=m;i++){
                int a;scanf("%d",&a);
                num[a]++;
            }
            for(int i=0;i<=b;i++) f[i]=1;
            for(int i=1;i<=n;i++){
                for(int j=b;j>=0;j--){
                    if(j>num[i]) 
                    f[j]-=f[j-num[i]-1],f[j]=(f[j]+mod)%mod;
                }
                for(int j=1;j<=b;j++) 
                f[j]+=f[j-1],f[j]%=mod;
            }
            printf("%d
    ",(f[b]-f[s-1]+mod)%mod);
        }
        return 0;
    }
  • 相关阅读:
    关于管理的经典故事(员工激励)
    通过SQL语句获取MSSQL数据库的数据字典
    AMF(java)学习笔记(一)
    ActionScript 3.0著名开源库 大集合
    Flash全屏功能详解
    NIO网络编程框架MINA2.0学习笔记(一)
    一份相当巨大的AS类库
    java NIO非阻塞式IO网络编程学习笔记(一)
    数据传输序列化与反序列化协议大全
    RED5学习笔记(一):RED5简介
  • 原文地址:https://www.cnblogs.com/plysc/p/11260778.html
Copyright © 2011-2022 走看看