Codeforces1183C(C题）Computer Game

发布时间：2020-12-16 10:49:21 所属栏目：百科来源：网络整理

导读：Vova is playing a computer game. There are in total? n n?turns in the game and Vova really wants to play all of them. The initial charge of his laptop battery (i.e. the charge before the start of the game) is? k k. During each turn Vova ca

Vova is playing a computer game. There are in total? $n$

During each turn Vova can choose what to do:

If the current charge of his laptop battery is strictly greater than? $a$
if the current charge of his laptop battery is strictly greater than? $b$
if the current charge of his laptop battery is less than or equal to? $a$

Regardless of Vova‘s turns the charge of the laptop battery is always decreases.

Vova wants to complete the game (Vova can complete the game if after each of? $n$

Your task is to find out the?maximum?possible number of turns Vova can?just play(make the?first type turn) or report that Vova cannot complete the game.

You have to answer? $q$

$q$

The first line of the input contains one integer? $q$

The only line of the query contains four integers? $k, n, a$

$k, a$

For each query print one integer:?-1?if Vova cannot complete the game or the?maximumnumber of turns Vova can?just play?(make the?first type turn) otherwise.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll q,k,a,b;
int main()
{
    cin>>q;
    while(q--)
    {
        cin>>k>>n>>a>>b;
        ll x;
        if((k-b*n)<=0) cout<<-1<<endl;
        else 
        {
            
            x=(k-n*b)/(a-b);if((k-n*b)%(a-b)==0)x--;
            cout<<min(x,n)<<endl;
        }
        
    }
    return 0;
}

思路分析：提供了电池剩余电量k,要使用轮数n,玩游戏消耗电量a,边充电边玩消耗电量b。先输入查询轮数q。然后输入每轮的k、n、a、b。如果电量减去所有轮数使用b方法消耗电量剩余电量还是小于等于0，就输出-1，代表没有使用a方法，且不能完成n轮任务。否则n轮全用b方法会有剩余电量，设变量x为总电量减去假设n轮全使用b方法消耗后剩余的电量再除以a、b两种电量使用方式之差，如果总电量减去假设n轮全使用b方法消耗后剩余的电量再除以a、b两种电量使用方式的余数为0.x减1.再输出x和n的最小值就是a方式再能完成n轮任务且有电量剩余情况下能使用的最多轮数。

 1 #include<iostream>
 2 using namespace std;
 3 int main() {
 4     long long q,cnt=-1;
 5     bool flag=false;
 6     cin>>q;
 7     long long k[q],n[q],a[q],b[q];
 8     for(long long i=0; i<q; i++) {
 9         cin>>k[i]>>n[i]>>a[i]>>b[i];
10     }
11     for(int i=0; i<q; i++) {
12         if(k[i]/a[i]>0) {
13             cnt=k[i]/a[i];
14         } else {
15             if(b[i]>=k[i]||b[i]*n[i]>=k[i]) {
16                 cout<<-1<<endl;
17             } else {
18                 cout<<0<<endl;
19             }
20         }
21         while(cnt>=0) {
22             if(cnt>n[i]) {// 9 17 18 15
23                 cout<<n[i]<<endl;
24                 break;
25             } else if(cnt<=n[i]) {
26                 if((n[i]-cnt)*b[i]+a[i]*cnt<k[i]) {
27                     cout<<cnt<<endl;
28                     flag==true;
29                     break;
30                 } else {
31                     cnt--;
32                 }
33                 if(cnt==0&&(n[i]-cnt)*b[i]>=k[i]) {
34                     cout<<-1<<endl;
35                 }
36             }
37         }
38         cnt=-1;
39     }
40 }

思路分析：先判断剩余电量k除以a方式消耗电量的值为正，再看最多可以使用几次a方式再加上剩余轮数全使用b方式的值要是小于剩余电量就输出使用a的次数cnt,否则将次数减一再进行同样判断。如果b方式消耗值大于剩余电量或者全部轮数使用b方式用电大于剩余电量k就将a的次数输出为-1，若全部轮数使用b方式用电小于等于剩余电量k就将a的次数输出为0。

（编辑：李大同）

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