POJ 3181 Dollar Dayz 完全背包 高精度dp 大数

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

题目大意：
给定k种硬币，每种硬币价值分别为1,2,3…k，每种硬币的数量不限，求出使用这k种硬币构成n元的方法

解题思路：
一看完题，瞬间想到了完全背包问题，然后想到了数组降维，动态规划方程式为

for (int i = 1; i <= k; i++) {//对每种硬币进行考虑
	//每一轮开始前，dp[j]是指只使用前i-1种硬币组成价值j的方法数
	for (int j = i; j <= n; j++) {
		dp[j] = dp[j] + dp[j - i];//dp[j-i]在这里是指用上了第i种硬币(至少一枚)
	}
	//每一轮结束后，dp[j]是指只使用前i种硬币组成价值j的方法数
}

得到代码为：

#include<iostream>
using namespace std;
int n, k;
int dp[1001];
int main() {
	while (cin >> n >> k) {
		dp[0] = 1;
		for (int i = 1; i <= 1001; i++) dp[i] = 0;
		for (int i = 1; i <= k; i++) {
			for (int j = i; j <=n; j++) {
				dp[j] += dp[j - i];
			}
		}
		//for (int i = 1; i <= n; i++) cout << dp[i] << endl;
		cout << dp[n] << endl;
	}
}

心情愉悦前去提交，然后得到了一发WA (ΩДΩ)
看了讨论区之后发现是要使用大数防止溢出，然后我又回去参考了书上关于高精度整数的那部分介绍，得到如下AC代码：
AC代码1

#include<iostream>
#include<stdio.h>
using namespace std;
int n, k;
struct bigInt {
	int digit[1001];
	int size;
	void init() {//初始化
		for (int i = 0; i <= 1000; i++) {
			digit[i] = 0;
		}
		size = 0;
	}
	void set(int x) {//用一个小整数设置高精度整数
		init();//初始化
		do {
			digit[size++] = x % 10000;
			x /= 10000;
		} while (x != 0);
	}
	void output() {//输出函数
		for (int i = size - 1; i >= 0; i--) {
			if (i == size - 1) { printf("%d", digit[i]); }
			else printf("%04d", digit[i]);
		}
		printf("\n");
	}
	bigInt operator + (const bigInt & A) const {
		bigInt ret;//保存结果
		ret.init();
		int carry = 0;//进位
		for (int i = 0; i < A.size || i < size; i++) {
			int tmp = A.digit[i] + digit[i] + carry;
			carry = tmp / 10000;
			tmp %= 10000;
			ret.digit[ret.size++] = tmp;
		}
		if (carry != 0) {
			ret.digit[ret.size++] = carry;
		}
		return ret;
	}
}dp[1001];//dp[i][j]从前i种价格中花j元
int main() {
	while (scanf("%d%d",&n,&k)!=EOF) {
		dp[0].set(1);
		for (int i = 1; i <= 1001; i++) dp[i].set(0);
		for (int i = 1; i <= k; i++) {
			for (int j = i; j <= n; j++) {
				dp[j] = dp[j] + dp[j - i];
			}
		}
		dp[n].output();
	}
}

写起来较为臃肿，且用时300+ms，然后我又看了一下大佬们的用时，基本都在50ms以内，看了一些大佬们的思路，发现可以用两个18位整数构成这个大数，我觉得挺不错的，嘿嘿
AC代码2

#include<iostream>
#include<stdio.h>
using namespace std;
int n, k;
long long dp1[1001];//高18位
long long dp2[1001];//低18位
const long long M = 1e18;
int main() {
	while (scanf("%d%d", &n, &k) != EOF) {
		dp2[0] = 1; dp1[0] = 0;
		for (int i = 1; i <= n; i++) {
			dp2[i] = 0; dp1[i] = 0;
		}
		long long carry = 0;
		for (int i = 1; i <= k; i++) {
			for (int j = i; j <= n; j++) {
				carry = (dp2[j] + dp2[j - i]) / M;
				dp1[j] = dp1[j] + dp1[j - i] + carry;
				dp2[j] = (dp2[j] + dp2[j - i]) % M;
			}
		}
		if (dp1[n] > 0) {
			printf("%lld%018lld\n", dp1[n], dp2[n]);
		}
		else printf("%lld\n", dp2[n]);
	}
}