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:
$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
题目意思：就是给出二个数N，和k，有1~k种钱币，每种都是无限个，用这些种类的钱币可以组合成总钱N有多少种方式。
解题：这就是一个完全背包，把N看成容量，钱币的类型值为花费和价值。与记录有多少种路径的走法一样，但要注意的是这里输出的数据会很大，会超出long long型，所以用两个long long 型来表示一个大数。分别是高位和低位，低位不超过10^18，也就是有17位数。
#include<stdio.h>
int N,dp1[1005];
__int64 dp[2][1005],c,mm;
void complexepack(int use)
{
for(int n=use;n<=N;n++)
if(dp1[n]<dp1[n-use]+use)
{
dp1[n]=dp1[n-use]+use;
dp[0][n]=dp[0][n-use];//低位
dp[1][n]=dp[1][n-use];//高位
}
else if(dp1[n]==dp1[n-use]+use)
{
c=(dp[0][n]+dp[0][n-use])/mm;//进位c
dp[0][n]=(dp[0][n]+dp[0][n-use])%mm;
dp[1][n]+=c+dp[1][n-use];
}
}
int main()
{
int k;
mm=1;
for(int i=1;i<=18;i++)mm*=10;
while(scanf("%d%d",&N,&k)>0)
{
for(int i=0;i<=N;i++)
{
dp[0][i]=dp[1][i]=0;
dp1[i]=0;
}
dp[0][0]=1;
for(int v=1;v<=k;v++)
complexepack(v);
if(dp[1][N])
printf("%I64d%017I64d\n",dp[1][N],dp[0][N]);
else
printf("%I64d\n",dp[0][N]);
}
}