描述

 http://poj.org/problem?id=3181

FJ有n元钱,有k种商品,各为1,2,...,k-1,k元,问有多少种花掉这n元钱的方法.

 

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5858		Accepted: 2197

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

Source

USACO 2006 January Silver

 

分析

 

这是一个完全部分和问题.对于多重部分和问题,可以用多重背包,但是由于有三层循环,会超时(即便使用二进制优化),所以有特殊的算法.而完全部分和问题可以直接用完全背包做.

注意:

1.是大数,要用高精度,或者两位分别存高低位:

unsigned long long 上限是1.8e19,所以每一位存1e17.

2.注意如果高位有数,低位要保证有前导0.

 

#include <cstdio>
#include <algorithm>
#define ull unsigned long long
const int maxw=1005,maxn=105;
const ull mod=100000000000000000;
int W,N;
ull dp[maxw][2];

void solve()
{
dp[0][0]=1;
for(int i=1;i<=N;i++)
{
for(int j=i;j<=W;j++)
{
dp[j][0]+=dp[j-i][0];
dp[j][1]+=dp[j-i][1];
dp[j][1]+=dp[j][0]/mod;
dp[j][0]%=mod;
}
}
if(dp[W][1])
{
printf("%llu",dp[W][1]);
printf("%017llu\n",dp[W][0]);
}
else
{
printf("%llu\n",dp[W][0]);
}
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("john.in","r",stdin);
freopen("john.out","w",stdout);
#endif
scanf("%d%d",&W,&N);
solve();
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
system("john.out");
#endif
return 0;
}

View Code