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Ant Counting (poj 3046 分組背包)

編輯：關於C++

Language: Ant Counting Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3142 Accepted: 1246

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

Sample Output

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

Source

USACO 2005 November Silver
題意：給出T個數字，編號1~T，每個數字有N[i]個。用這些數字組成一個序列，最後問長度在S~B之間的不同序列有多少種。

思路：分組背包。每個螞蟻家族為一個分組，在每個分組裡的選擇有N[i]種（選1~N[i]）。背包九講

代碼：

#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

int dp[maxn*100];
int num[maxn];
int t,a,s,b;

int main()
{
    int i,j,x,k;
    while (scanf("%d%d%d%d",&t,&a,&s,&b)!=EOF)
    {
        mem(num,0);
        mem(dp,0);
        FRE(i,1,a)
        {
            sf(x);
            num[x]++;
        }
        dp[0]=1;
        FRE(i,1,t)
        {
            FREE(j,a,0)
            {
                FRE(k,1,num[i])
                {
                    if (j-k>=0)
                    {
                        dp[j]=(dp[j]+dp[j-k])%1000000;
                    }
                }
            }
        }
        int ans=0;
        FRE(i,s,b)
            ans=(ans+dp[i])%1000000;
        pf("%d\n",ans);
    }
    return 0;
}