poj3046

Ant Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6012 Accepted: 2240

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 312213

Sample Output

10

题意：有A个数，一共T种，从中选一些数作为一个集合，问集合大小在A到B中的种类有多少

思路：相同的数我们可以直接一起处理，dp[i+1][j]表示从前i种物品数里取出j个的组合数，那么可以得出dp[i][j]=Σdp[i-1][j-k]，优化一下就成了 dp[i][j] = dp[i-1][j] + dp[i][j-1] - dp[i-1][j-ant[i]-1]。

代码：

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <map>using namespace std;#define ma(a) memset(a,0,sizeof(a))const long long int MOD=1000000;long long int num[1005];long long int dp[1005][100005];int main(){    int T,A,S,B;    while(scanf("%d %d %d %d",&T,&A,&S,&B)!=EOF)    {        int i,j,k;        ma(num);        for(i=0;i<A;i++)        {            int x;            scanf("%d",&x);            num[x]++;        }        int maxn=max(S,B);        for(i=1;i<=T+1;i++)            dp[i][0]=1;        for(i=1;i<=T;i++)        {            for(j=1;j<=maxn;j++)            {                if(j-1-num[i]>=0)                {                    dp[i+1][j]=(dp[i+1][j-1]+dp[i][j]-dp[i][j-1-num[i]]+MOD)%MOD;                }                else                {                    dp[i+1][j]=(dp[i+1][j-1]+dp[i][j])%MOD;                }            }        }        long long int ans=0;        for(i=S;i<=B;i++)        {            ans=(ans+dp[T+1][i])%MOD;        }        cout<<ans<<endl;    }    return 0;}