我们可以o(n^2),枚举每一个布告,然后从后往前枚举i前面的位置,然后状态转移
void solve() {
int n;
cin >> n;
vector<int> a(n + 1);
for (int i = 1; i <= n; i++) cin >> a[i];
vector<int> f(n + 1, 0x3f3f3f3f);
f[0] = 0;
for (int i = 1; i <= n; i++) {
for (int j = i - 1; j >= 0; j--) {
if (j + a[i] >= i) {
if (j + a[i] <= n) f[j + a[i]] = min(f[j + a[i]], f[j] + 1);
}
}
}
if (f[n] >= 0x3f3f3f3f) f[n] = -1;
cout << f[n] << endl;
}