codeforces 1141E 模拟+math？

xg

题意

　　给你n个数，和H，第pos次，sumpos大于H（sum为前缀和）。求最小的pos，pos属于1到+无穷。输出最小的pos，无解输出-1。

思路

对于1-n。如果sum>=H，则直接输出。

对于大于n。如果sum.n小于H，则无解。

　　　　　　否则，则答案一定为 i+k*n（i为【1，n】，k为【2，+无穷】。

　　　　　　则，遍历1-n，求出对于每一个位置最小的k。然后更新ans即可。

　　　　　　k = （H-sum【i】）/sum + ((H-b[i])%sum != 0)

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '\n'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
ll a[maxn];
ll b[maxn];
int main()
{
    //freopen("input.txt", "r", stdin);
    ll H, n;
    cin>>H>>n;
    ll minn = 0;
    ll sum = 0;
    rep(i,1,n){
        cin>>a[i];
    }
    rep(i,1,n){
        sum += a[i];
        b[i] = -sum;
        if(sum < 0 && abs(sum) >= H){
            cout<<i<<endl;
            return 0;
        }
    }
    if(sum >= 0) {
        cout<<-1<<endl;
        return 0;
    }
    sum = -sum;
    ll ans = 9e18;
    for(ll i = 1; i <= n;++i){
        ll t = H-b[i];
        ll cnt = (H-b[i])/sum;
        if((H-b[i])%sum != 0) cnt++;
        ll tmp = cnt*n + i;
        ans = min(ans,tmp); 
    }
    cout<<ans<<endl;


}