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

E. Superhero Battle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 kk such 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




这个题目不难，但是细节比较多，我觉得我做的很自闭，因为WA好多次，然后我就自闭了。
上网查了查题解，就是要考虑一次循环之后可以减少最多的值，这个要优先考虑，然后就是去找有多少个整次的循环。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 100;
ll a[maxn];

int main()
{
    int flag=0,mark;
    ll h, n,sum=0,mi=inf;
    cin >> h >> n;
    ll g = h;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld", &a[i]);
        sum += a[i];
        mi = min(mi, sum);
        g += a[i];
        if(g<=0&&!flag)
        {
            mark = i;
            flag = 1;
        }
    }
    if(flag)
    {
        printf("%d
", mark);
        return 0;
    }
    if (sum >= 0)
    {
        printf("-1
");
        return 0;
    }
    sum = -sum;
    h += mi;
    ll ans = h / sum * n;
    ll cur = h % sum - mi;
    while(cur>0)
    {
        for(int i=1;i<=n;i++)
        {
            cur += a[i];
            ans++;
            if (cur <= 0) break;
        }
        //printf("cur=%d
", cur);
    }
    printf("%lld
", ans);
    return 0;
}