[POJ3181] Dollar Dayz-优快云博客

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.

题解：

dp+高精加

显然完全背包dp，dp[j]+=dp[j-i]

需要高精加法

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;

const int N = 1010;

int dp[N][110];

int gi() {
  int x=0,o=1; char ch=getchar();
  while(ch!='-' && (ch<'0' || ch>'9')) ch=getchar();
  if(ch=='-') o=-1,ch=getchar();
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return o*x;
}

void plu(int x, int y) {
  int c=0,k=max(dp[x][0],dp[y][0]);
  for(int i=1; i<=k; i++) {
    dp[x][i]=dp[x][i]+dp[y][i]+c;
    c=dp[x][i]/10;
    dp[x][i]%=10;
  }
  dp[x][0]=k;
  while(c) {
    dp[x][++dp[x][0]]=c%10;
    c/=10;
  }
}

int main() {
  int m=gi(),n=gi();
  dp[0][0]=1,dp[0][1]=1;
  for(int i=1; i<=n; i++)
    for(int j=i; j<=m; j++)
      plu(j,j-i);
  for(int i=dp[m][0]; i>=1; i--)
    printf("%d", dp[m][i]);
  return 0;
}

转载于:https://www.cnblogs.com/HLXZZ/p/7509472.html