纸币问题 2

这是我的纸币问题 2 的代码

#include <iostream>
using namespace std;
const int S = 1e3 + 5, N = 1e4 + 5, I = 1 << 30, M = 1e9 + 7;
int n, w, a[S], f[N];
void read(int &a) {
	int s = 0, f = 1; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') s = s * 10 + c - 48, c = getchar();
	a = s * f;
} 
int main() {
	read(n), read(w);
	for (int i = 1; i <= n; i++) read(a[i]);
	f[0] = 1;
	for (int i = 1; i <= w; i++)
		for (int j = 1; j <= n; j++) 
			if (i - a[j] >= 0) f[i] = (f[i] + f[i - a[j]]) % M;
	printf("%d", f[w]);
	return 0;
}

这是纸币问题 3 的代码

#include <iostream>
using namespace std;
const int S = 1e3 + 5, N = 1e4 + 5, I = 1 << 30, M = 1e9 + 7;
int n, w, a[S], f[N];
void read(int &a) {
	int s = 0, f = 1; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') s = s * 10 + c - 48, c = getchar();
	a = s * f;
} 
int main() {
	read(n), read(w);
	for (int i = 1; i <= n; i++) read(a[i]);
	f[0] = 1;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= w; j++) 
			if (j - a[i] >= 0) f[j] = (f[j] + f[j - a[i]]) % M;
	printf("%d", f[w]);
	return 0;
}

这两个代码只有枚举 $n$ 和 $w$ 的先后的区别，我之所以这么写确实是我感觉两者有些区别，应该就是 lyd 蓝书里“阶段”和“转移”的区别（记错了勿喷 Orz），但我细想感觉状态的定义十分类似，都是 $f_i$ 为凑齐 $i$ 元的方案数/组合数，我感觉有些区别但是又说不清楚问什么，求大佬教教为啥枚举 $n$ 和 $w$ 的顺序不一样 Orz