POJ 3181 Dollar Dayz （dp）

Dollar Dayz

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4442 Accepted: 1716

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

划分数 dp[i][j] = dp[i-1][j] + dp[i][j-i];

- - 要高精 然后dph高位 dpl低位 滚动数组优化

AC代码如下：

////  POJ 3181 Dollar Dayz////  Created by TaoSama on 2015-03-05//  Copyright (c) 2015 TaoSama. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>#define CLR(x,y) memset(x, y, sizeof(x))using namespace std;const int INF = 0x3f3f3f3f;const long long MOD = 1e18;const int N = 1e5 + 10;int n, k;long long dph[N], dpl[N]; //¸ßλµÍλint main() {#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifios_base::sync_with_stdio(0);while(cin >> n >> k) {memset(dph, 0, sizeof dph);memset(dpl, 0, sizeof dpl);dpl[0] = 1;for(int i = 1; i <= k; ++i) {for(int j = i; j <= n; ++j) {dph[j] = dph[j] + dph[j - i] + (dpl[j] + dpl[j - i]) / MOD;dpl[j] = (dpl[j] + dpl[j - i]) % MOD;}}if(!dph[n]) cout<<dpl[n]<<endl;elsecout<<dph[n]<<setw(18)<<setfill('0')<<dpl[n]<<endl;}return 0;}