C 解法, 执行用时: 2ms, 内存消耗: 340KB, 提交时间: 2021-10-19

#include <stdio.h>

int main()
{
    int HP, normalAttack, buffedAttack;
    int count = 0;
    scanf("%d\n%d\n%d", &HP, &normalAttack, &buffedAttack);
    if(HP < normalAttack)
        count = 1;
    else if(HP < buffedAttack)
        count = 2;
    else if(2*normalAttack >= buffedAttack)
    {
        count = HP/normalAttack;
        if(HP%normalAttack)
            count++;
    }
    else
    {
        count = HP/buffedAttack*2;
        int rem = HP%buffedAttack;
        if(rem != 0 && rem <= normalAttack)
            count++;
        else if(rem > normalAttack)
            count+=2;
    }
    printf("%d", count);
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 372KB, 提交时间: 2021-10-07

#include<stdio.h>

int up(int n, int m) {
    if(n % m == 0) {
        return n/m;
    }
    else {
        return n/m + 1;
    }
}

int main() {
    int hp,normalAttack,buffedAttack;
    scanf("%d",&hp);
    scanf("%d",&normalAttack);
    scanf("%d",&buffedAttack);
    if(normalAttack * 2 >= buffedAttack) {
        printf("%d",up(hp,normalAttack));
    }
    else {
        int temp = hp % buffedAttack;
        if(temp <= normalAttack && temp != 0) {
            printf("%d",2*hp/buffedAttack+1);
        }
        else {
            printf("%d",2*up(hp,buffedAttack));
        }
    }
}

C 解法, 执行用时: 2ms, 内存消耗: 372KB, 提交时间: 2020-09-09

#include <stdio.h>
int main()
{
    int HP,normalAttack,buffedAttack,num;
    scanf("%d",&HP);
    scanf("%d",&normalAttack);
    scanf("%d",&buffedAttack);
    if(normalAttack*2>buffedAttack)
    {
        num=HP/normalAttack;
        if(HP%normalAttack!=0)
        {
            num++;
        }
    }
    else
    {
        num=HP/buffedAttack * 2;
        if(HP%buffedAttack!=0)
        {
            if(HP%buffedAttack>normalAttack)
            {
                num+=2;
            }
            else
            {
                num++;
            }
        }
    }
    printf("%d\n",num);
    return 0;
}

/*#include <stdio.h>
int main()
{
    int HP,buff,norm;
    int i=0,j=0,k=0,p=0;
    scanf("%d",&HP);
    scanf("%d",&norm);
    scanf("%d",&buff);
    double n1=HP,n2=HP,n3=HP,n4=HP;
    while(n1>0)
    {
        n1=n1-norm;
        i++;
    }
    while(n2>0)
    {
        n2=n2-buff;
        j+=2;;
    }
    while(n3>0)
    {
        n3=n3-buff;
        k+=2;;
        n3=n3-norm;
        k++;
    }
    while(n4>0)
    {
        n4=n4-norm;
        p++;
        n4=n4-buff;
        p+=2;;
    }
    int a[4]={i,j,k,p};
    for(int w=0;w<3;w++)
    {
        for(int e=0;e<4-1-w;e++)
        {
            if(a[e+1]<a[e])
            {
                int temp=a[e+1];
                a[e+1]=a[e];
                a[e]=temp;
            }
        }
    }
    printf("%d\n",a[0]);
    return 0;
}*/

C++14 解法, 执行用时: 2ms, 内存消耗: 372KB, 提交时间: 2020-08-03

#include <cstdio>
#include <cmath>
int main(){
    long long s, b, n;
    scanf("%lld%lld%lld", &s, &n, &b);
    if (2*n < b){
        int t = s / b;
        int r = s % b;
        if (!r) printf("%lld\n", 2*t);
        else {
                    if (r <= n) printf("%lld\n", 2*t + 1);
        else printf("%lld\n", 2*t + 2);
        }

    }
    else{
        printf("%lld\n", (long long)(ceil(s*1.0/n)));
    }
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 372KB, 提交时间: 2020-05-01

#include<stdio.h>
main()
{
    int hp;
    while(scanf("%d",&hp)!=EOF)
    {
        int normal,buff,min,k,remain;
        scanf("%d\n%d",&normal,&buff);
        if (buff<=2*normal)
        {
            min=hp/normal;
            remain=hp-min*normal;
            if(remain==0) min=min;
            else min=min+1;
        }
        else
        {
            k=hp/buff;
            min=k*2;
            remain=hp-k*buff;
            if (remain==0) min=min;
            else if(remain<=normal) min=min+1;
            else min=min+2;
        }
        printf("%d",min);
    }
}