1752:鸡兔同笼（2.1）

最朴素的方法：

#include<cstdio>
int main(){
    int a,maxn=-1,minn=32769;
    scanf("%d",&a);
    if (a%2==0) maxn=a/2;
    if (a%4==0) minn=a/4;
    for (int t=1;t<=a/4;t++)
        for(int j=1;j<=a/2;j++)
            if(4*t+2*j==a){
                if (t+j<minn) minn=t+j;
                if (t+j>maxn) maxn=t+j;
            }
    if (maxn==-1&&minn==32769) printf("0 0\n");
    else     printf("%d %d\n",minn,maxn);
    return 0;
}

时间：32768/4*32768/2=134217728~1*10⁸

如果测试数据过大未必能通过

简单优化：

#include<cstdio>
int main(){
    int a,maxn=-1,minn=32769;
    scanf("%d",&a);
    if (a%2==0) maxn=a/2;
    if (a%4==0) minn=a/4;
    if (a%2>0){
        printf("0 0\n");
        return 0;
    }
    if (maxn==-1||minn==32769) {
    for (int t=a/4;t>=0;t--)
        for(int j=1;j<=a/2;j++)
            if(4*t+2*j==a){
                if (t+j<minn) minn=t+j;
                if (t+j>maxn) maxn=t+j;
            }
    }
    if (maxn==-1&&minn==32769) printf("0 0\n");
    else     printf("%d %d\n",minn,maxn);
    return 0;
}

继续优化：保留一重循环

如果有解则a定为偶数，则最大值定为a/2。如果a%4==0则最小值为 minn=a/4;

//如果有解则a定为偶数，则最大值定为a/2。如果a%4==0则最小值为 minn=a/4;
#include<cstdio>
int main(){
    int a,maxn=-1,minn=32769;
    scanf("%d",&a);
    if (a%2==0) maxn=a/2;
    else {//如果a非偶肯定无解
        printf("0 0\n");
        return 0;
    }
    if (a%4==0) minn=a/4;
    if (minn==32769)
        for (int t=1;t<=a/4;t++){
            int j=(a-4*t)/2;
            if (t+j<minn) minn=t+j;
        }
//    if (maxn==-1&&minn==32769) printf("0 0\n");
    printf("%d %d\n",minn,maxn);
    return 0;
}

三种方法的时间复杂度：