(完全背包 大数）Dollar Dayz （POJ 3181）

Posted 2020-07-15 栀蓝

http://poj.org/problem?id=3181

 

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

一看完全背包， 套了模板， 很遗憾wa了， 想了想，不明所以， 搜了下题解， 居然说是因为结果太大， 看看数据似乎是这样的， 据说是两位数存就可以， 结果写完提交，果断AC

 

#include <iostream>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <stack>
#include <vector>
#include <algorithm>

using namespace std;

#define N 2100
#define met(a,b) (memset(a,b,sizeof(a)))
typedef long long LL;


LL dp[N][2];

int main()
{
    int n, m;

    while(scanf("%d%d", &n, &m)!=EOF)
    {
        int i, j;

        met(dp, 0);

        dp[0][0] = 1;
        for(i=1; i<=m; i++)
        for(j=i; j<=n; j++)
        {
            dp[j][0] += dp[j-i][0];
            dp[j][1] += dp[j-i][1];
            dp[j][1] += dp[j][0]/1000000000000000;
            dp[j][0] %= 1000000000000000;
        }

        if(dp[n][1]==0)
            printf("%I64d\\n", dp[n][0]);
        else
        {
            printf("%I64d%015I64d\\n", dp[n][1], dp[n][0]);
        }
    }

    return 0;
}

 

 

参考http://www.cnblogs.com/kuangbin/archive/2012/09/20/2695165.html