转移的时候枚举一下当前的$i$个人干什么。
对于$sta$需要分类讨论，代码里有注释
```cpp
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, x, y, z, A, B;
int f[41][21][21][4];
void Min(int &x, int y) {
if(y < x) x = y;
}
main() {
#ifdef WIN32
freopen("a.in", "r", stdin);
#endif
N = read(); x = read(); y = read(); z = read(); A = read(); B = read();
int ans = INF;
memset(f, 0xf, sizeof(f));
f[N][0][0][0] = 0;
for(int i = N; i <= 40; i++) {
for(int sta = 0; sta <= 3; sta++) {
for(int a = 0; a <= max(A, z); a++) {//已经有了a的金钱
for(int b = 0; b <= B; b++) {//b的荣誉
if(f[i][a][b][sta] < ans) {
int cur = f[i][a][b][sta];
if(a >= A && b >= B) {ans = min(ans, cur); continue;}
for(int pep = 0; pep <= i; pep++) {//有i个人挣钱
int na = min(a + pep * x, max(A, z)), nb = min(B, b + y * (i - pep));
if(sta == 0) {
Min(f[i][na][nb][0], cur + 1);
if(na >= z) Min(f[i][na - z][nb][1], cur + 1);
}
//当前没有发广告，接下来可以不发广告，可以发广告
else if(sta == 1 || sta == 2) Min(f[i][na][nb][sta + 1], cur + 1);
//当前已经发了广告，只能等着。。
else {
Min(f[i + 1][na][nb][0], cur + 1);
if(na >= z) Min(f[i + 1][na - z][nb][1], cur + 1);
}
//当前发完了广告，考虑是否继续发广告
}
}
// if(f[i][a][b][sta] <= 100000)printf("%d %d %d %d %d\n", i, a, b, sta, f[i][a][b][sta]);
}
}
}
}
printf("%d", ans);
}
```