Codeforces Round #547 (Div. 3) E. Superhero Battle

E. Superhero Battle

A superhero fights with a monster. The battle consists of rounds, each of which lasts exactly 𝑛n minutes. After a round ends, the next round starts immediately. This is repeated over and over again.

Each round has the same scenario. It is described by a sequence of 𝑛n numbers: 𝑑1,𝑑2,…,𝑑𝑛d1,d2,…,dn (−106≤𝑑𝑖≤106−106≤di≤106). The 𝑖i-th element means that monster's hp (hit points) changes by the value 𝑑𝑖di during the 𝑖i-th minute of each round. Formally, if before the 𝑖i-th minute of a round the monster's hp is ℎh, then after the 𝑖i-th minute it changes to ℎ:=ℎ+𝑑𝑖h:=h+di.

The monster's initial hp is 𝐻H. It means that before the battle the monster has 𝐻H hit points. Print the first minute after which the monster dies. The monster dies if its hp is less than or equal to 00. Print -1 if the battle continues infinitely.

Input

The first line contains two integers 𝐻H and 𝑛n (1≤𝐻≤10121≤H≤1012, 1≤𝑛≤2⋅1051≤n≤2⋅105). The second line contains the sequence of integers 𝑑1,𝑑2,…,𝑑𝑛d1,d2,…,dn (−106≤𝑑𝑖≤106−106≤di≤106), where 𝑑𝑖di is the value to change monster's hp in the 𝑖i-th minute of a round.

Output

Print -1 if the superhero can't kill the monster and the battle will last infinitely. Otherwise, print the positive integer 𝑘ksuch that 𝑘k is the first minute after which the monster is dead.

Examples

input

Copy

1000 6
-100 -200 -300 125 77 -4

output

Copy

9

input

Copy

1000000000000 5
-1 0 0 0 0

output

Copy

4999999999996

input

Copy

10 4
-3 -6 5 4

output

Copy

-1

题意：给定一个怪的hp，和一轮的攻击次数n；然后给n次攻击的伤害（有正有负），如果一轮内怪不死会一直按顺序重复这n次的攻击，问攻击到多少下怪会死。

思路：二分答案，但是在二分之前要特判一下怪打不死之类的情况；用一个前缀和记录一轮中直至目前造成的伤害和，二分内部每次判这一轮能不能把怪打死，如果能，更新一下攻击次数。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define ll long long
#define lowbit(x) (x&(-x))
#define PI acos(-1)
#define ms(x,y) memset(x, y, sizeof(x))
using namespace std;

const int maxn = 2e5+7;
ll arr[maxn];
ll pre[maxn];

int main()
{
    ll hp, tot = 0, maxx = 0;
    int n;
    scanf("%lld%d", &hp, &n);
    for(int i=0;i<n;i++)
    {
        scanf("%lld", arr+i);
        tot += arr[i];
        pre[i] = i ? pre[i-1] + arr[i] : arr[i];
        if(-pre[i] > maxx) maxx = -pre[i];
    }
    
    if(maxx < hp && tot >= 0)
    {
        printf("-1\n"); // never die
        return 0;
    }
    else if(maxx >= hp && tot >= 0) // die on round 1
    {
        for(int i=0;i<n;i++)
            if(-pre[i] >= hp)
            {
                printf("%d\n", i+1); break;
            }
        return 0;
    }
    else // must die
    {
        tot = -tot;
        ll l = 1, r = hp/tot + (hp%tot != 0) + 1, mid;
        ll minn = r;
        ll ans = 0;
        while(l < r)
        {
            mid = l + (r-l) / 2; // this round
            ll tp = hp - tot * (mid-1); // have deducted before
            if(maxx >= tp) // will die this round
            {
                ll tp_ans = ans;
                for(int i=0;i<n;i++)
                    if(-pre[i] >= tp)
                    {
                        tp_ans = i+1 + (mid-1)*n;
                        break;
                    }
                if(mid < minn)
                {
                    minn = mid;
                    ans = tp_ans;
                }
                r = mid;
            }
            else l = mid+1;
            
        }
        printf("%lld\n", ans);
    }
}