POJ 3181 Dollar Dayz(dp,背包）

dp[i][j] 表示 用 前i种前能 凑成j 元的方法数。

dp[i][j] = dp[i-1][j] + dp[i][j- (i+1)]  因为我i是从0开始的。。 所以i+1 表示第i中的值

最后要用一下高精度

题目：

H - Dollar Dayz

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description:

Description

Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $1, $2, and $3. Farmer John has exactly $5 to spend. He can buy 5 tools at $1 each or 1 tool at $3 and an additional 1 tool at $2. Of course, there are other combinations for a total of 5 different ways FJ can spend all his money on tools. Here they are:

        1 @ US$3 + 1 @ US$2

        1 @ US$3 + 2 @ US$1

        1 @ US$2 + 3 @ US$1

        2 @ US$2 + 1 @ US$1

        5 @ US$1

Write a program than will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of $1..$K (1 <= K <= 100).

Input

A single line with two space-separated integers: N and K.

Output

A single line with a single integer that is the number of unique ways FJ can spend his money.

Sample Input

5 3

Sample Output

5


代码：

#include <iostream>
#include <cstdio>
using namespace std;

int N,K;
int dp[1000+10][1000+10];

void epl(int a,int b)
{
    for(int i=0;i<=60;i++)
    {
        dp[a][i] = dp[b][i];
    }
}

void add(int a,int b)
{
    for(int i=0;i<60;i++)
    {
        dp[a][i]+= dp[b][i];
        if(dp[a][i]>=10)
        {
            dp[a][i]%=10;
            dp[a][i+1] ++;
        }
    }
}

int main()
{
    cin>>K>>N;
    dp[0][0]=1;
    for(int i=0;i<N;i++)
    {
        for(int k=0;k<=K;k++)
        {
            //dp[i+1][k] = dp[i][k];
            if( k>=i+1)
            {
                //dp[i+1][k]+= dp[i+1][k-i-1];
                add(k,k-i-1);
            }
        }
    }
    int t = 60;
    while(t>0 && dp[K][t]==0)t--;
    for(int i=t;i>=0;i--)
    {
        printf("%d",dp[K][i]);
    }
    //cout<<dp[N][K];
    printf("
");
    return 0;
}