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：求1~m中的整数有多少种组成n的方法。

方法1：完全背包问题，注意高精度即可。

方法2：我用的是动态规划来拆分整数，想起来更容易些。设dp[n][m]表示用不超过m的整数表示n的方法数，分情况讨论：

m=0时，dp[n][0]=0；

m>n时，dp[n][m]=dp[n][n]；

0<m<=n时，若拆分出的整数最大是m，则方法数为dp[n-m][m]，若最大整数小于m，则方法数为dp[n][m-1]。所以总数为dp[n][m]=dp[n-m][m]+dp[n][m-1]。

空间还可以进一步优化，利用滚动数组，减去第二维。

最后就是注意高精度，将大数分割成两部分，右半部分属于long long范围内，左半部分除以1e18后另外保存，具体见代码。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long LL;
const LL MOD = 1e18;

LL a[1005], b[1005];

int main()
{
    int n, m;
    cin >> n >> m;
    a[0] = 1;
    for (int j = 1; j <= m; ++j)
        for (int i = j; i <= n; ++i)
        {
            b[i] = b[i-j] + b[i] + (a[i-j] + a[i]) / MOD;
            a[i] = (a[i-j] + a[i]) % MOD;
        }
    if (b[n])
        printf("%lld", b[n]);
    printf("%lld
", a[n]);
    return 0;
}