题解：
先求出最后的单价，我们每一步都于逼近单价。
因为只有总价部分可以优化成加一个整数，所以每一步去就加上目标总价与当前总价差多少即可。
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <stack>
#include <cmath>
#include <deque>
#include <queue>
#include <list>
#include <set>
#include <map>
#define line printf("---------------------------\n")
#define mem(a, b) memset(a, b, sizeof(a))
#define pi acos(-1)
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 2000+10;
int main(){
int x, y;
while(~scanf("%d %d", &x, &y)){
if(y < x){
printf("-1\n");
continue;
}
double sum = 1.0, univalence = (y+1.0-0.01)/x;
int ans = x-1;
for(int i = 1; i <= x; i++){
int k = (int)(univalence*i-sum);
ans += k;
sum += k;
sum = sum*(i+1.0)/i;
}
printf("%d\n", ans);
}
}