POJ-3181 Dollar Dayz（划分数+高精度）

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

---

#include <iostream>
#include <vector>
#include <cmath>
#include <string>
#include <algorithm>

using namespace std;


int n = 105, m = 1005;
vector<vector<long long> > before(n, vector<long long>(m)); //前18位数
vector<vector<long long> > after(n, vector<long long>(m)); //后18位数
long long M = 1e18;

int main()
{
    std::ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    after[0][0] = 1;

    int m, n;
    while (cin >> m >> n) {
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j <= m; ++j) {
                if (j >= i) {
                    before[i][j] = before[i - 1][j] + before[i][j - i] + (after[i - 1][j] + after[i][j - i]) / M;
                    after[i][j] = (after[i - 1][j] + after[i][j - i]) % M;
                }
                else {
                    before[i][j] = before[i - 1][j];
                    after[i][j] = after[i - 1][j];
                }
            }
        }

        if (before[n][m]) cout << before[n][m];
        cout << after[n][m] << endl;
    }

    return 0;
}

这道题目其实就是划分数的模板题，但是因为数据范围导致结果很大，超过了long long的标识范围，所以这个题目一个很巧妙的地方是可以用两个long long的数组合起来表示一个大整数。用数组before表示前18位的数字，after表示后18位的数字，用一个大数M来考察如何划分。