poj3181 Dollar Dayz

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

题意：给你两个数n。k，让你用1到k这k个数表示n，问有几种方法，本质是整数的拆分。由于最后结果比較大，超过long long ,所以用两个long long连接起来，设两个数组a[][],b[][]分别表示没有超过long long的部分以及超过long long 部分。

当中a[n][m]表示n用一些数拆分，当中最大的数不超过m的方案数，绘图能够看到。当n<m时，a[i][j]=a[i][i];当n>=m时。a[i][j]=(a[i][j-1]+a[i-j][j])%inf;初始化的时候要令a[0][i]=1,由于a[2][2]=a[0][2]+a[2][1];

6

5 + 1

4 + 2, 4 + 1 + 1

3 + 3, 3 + 2 + 1, 3 + 1 + 1 + 1

2 + 2 + 2, 2 + 2 + 1 + 1, 2 + 1 + 1 + 1 + 1

1 + 1 + 1 + 1 + 1 + 1

#include<stdio.h>
#include<string.h>
#define ll long long
ll a[1006][106],b[1006][106];
ll inf;
int main()
{
int n,m,i,j;
inf=1;
for(i=1;i<=18;i++){
inf*=10;
}
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(i=1;i<=m;i++){
a[0][i]=1;
}
for(i=1;i<=n;i++){
a[i][1]=1;
}
for(j=1;j<=m;j++){
a[1][j]=1;
}

for(j=2;j<=m;j++){
for(i=2;i<=n;i++){
if(i<j){
a[i][j]=a[i][i];
b[i][j]=b[i][i];
}
else{
a[i][j]=(a[i][j-1]+a[i-j][j])%inf;
b[i][j]=b[i][j-1]+b[i-j][j]+(a[i][j-1]+a[i-j][j])/inf;
}
//printf("%d %d %d\n",i,j,a[i][j]);
}
}
if(b[n][m])
printf("%lld",b[n][m]);
printf("%lld\n",a[n][m]);
}
return 0;
}

这题也能够用全然背包，并用高精度模拟。状态转移方程：dp[j]+=dp[j-w[i]]

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define inf 0x7fffffff
#define ll long long
int dp[1005][60];
void add(int a,int b){
    int i,j;
    for(i=1;i<=50;i++){
        if(dp[a][i]+dp[b][i]<=9){
            dp[a][i]=dp[a][i]+dp[b][i];
        }
        else{
            dp[a][i]=(dp[a][i]+dp[b][i])%10;
            dp[a][i+1]++;
        }
    }
}

int main()
{
    int n,m,i,j,k,t;
    while(scanf("%d%d",&m,&k)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        dp[0][1]=1;
        for(i=1;i<=k;i++){
            for(j=i;j<=m;j++){
                add(j,j-i);
            }
        }
        t=50;
        while(t>=2 && dp[m][t]==0)t--;

        for(i=t;i>=1;i--){
            printf("%d",dp[m][i]);
        }
        printf("\n");
    }
    return 0;
}