time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
A superhero fights with a monster. The battle consists of rounds, each of which lasts exactly n 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 n numbers: d1,d2,…,dn (−106≤di≤106). The i-th element means that monster’s hp (hit points) changes by the value di during the i-th minute of each round. Formally, if before the i-th minute of a round the monster’s hp is h, then after the i-th minute it changes to h:=h+di.
The monster’s initial hp is H. It means that before the battle the monster has H hit points. Print the first minute after which the monster dies. The monster dies if its hp is less than or equal to 0. Print -1 if the battle continues infinitely.
Input
The first line contains two integers H and n (1≤H≤1012, 1≤n≤2⋅105). The second line contains the sequence of integers d1,d2,…,dn (−106≤di≤106), where di is the value to change monster’s hp in the i-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 k such that k is the first minute after which the monster is dead.
Examples
input
1000 6
-100 -200 -300 125 77 -4
output
9
input
1000000000000 5
-1 0 0 0 0
output
4999999999996
input
10 4
-3 -6 5 4
output
-1
题意：一直怪兽有血量H，给一个序列表示每轮怪兽每次收到的伤害和回复的血量，现在判断怪兽要经历多次操作会死亡。
H值很大，暴力模拟是不可能了，因此做一些小优化，如果怪兽的血量能撑过每一轮攻击的最大伤害值，那么说明至少在这一轮他是不会死的，因此对于一个不会死的轮数，直接除过去就好了，求一个sum表示每轮遭到的伤害，求一个max表示每轮伤害的巅峰，能承受的一轮，减去sum，不能承受的一轮，即最后一轮，直接模拟怪兽收到的每次伤害即可。
用（H-max）/ sum即怪兽能完整撑过的轮数，用轮数乘每轮操作次数即可得到操作总数，最后模拟每次操作，减到怪物H为0即可
代码如下：
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn=2e5+7;
int n;
LL hp,a[maxn];
int main()
{
LL sum=0,mx=0;
bool flag=false;
scanf("%lld%d",&hp,&n);
for(int i=0; i<n; i++)
{
scanf("%lld",&a[i]);
sum+=a[i];
if(-sum>=hp)flag=true;
if(-sum>mx)mx=-sum;
}
if(sum>=0&&!flag)printf("-1
");
else
{
LL cnt=0;
if(flag)
{
int i=0;
while(hp>0)hp+=a[i++],cnt++;
printf("%lld
",cnt);
return 0;
}
cnt=(hp-mx)/abs(sum);
hp-=cnt*abs(sum);
if(!hp)hp=abs(sum);
cnt*=n;
int i=0;
while(hp>0) hp+=a[i++],cnt++,i=i==n?0:i;
printf("%lld
",cnt);
}
}