codeforces#512 Div2

pre过了三题 终测又挂了一题 又掉分了 真的是 太菜了

 

A-In Search of an Easy Problem

水题 有一个1就是hard

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;

int n;

int main()
{
    while(scanf("%d", &n) != EOF){
        int easy;
        bool flag = false;
        for(int i = 0; i < n; i++){
            scanf("%d", &easy);
            if(easy == 1){
                flag = true;
            }
        }
        if(flag){
            printf("HARD\n");
        }
        else{
            printf("EASY\n");
        }
    }
    return 0;
}

 

 

B-Vasya and Cornfield

求一下四条边的直线方程 代入点判断即可

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;

int n, d, m;
struct node{
    int x, y;
};

int main()
{
    while(scanf("%d%d%d", &n, &d, &m) != EOF){
        for(int i = 0; i < m; i++){
            int x, y;
            scanf("%d%d", &x, &y);
            if(y >= -x + d && y <= -x + 2 * n - d && y >= x - d && y <= x + d){
                printf("YES\n");
            }
            else{
                printf("NO\n");
            }
        }
    }
    return 0;
}

 

C-Vasya and Golden Ticket

终测挂了的题

求前缀和 对于每一个i看sum[n]是sum[i]的几倍 然后看i之后有没有1-k倍的sum[i]

挂了是因为没判断每个倍数都有 找到一个就输出了

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;

int n;
char s[105];
int presum[105];

int main()
{
    while(scanf("%d", &n) != EOF){
        memset(presum, 0, sizeof(presum));
        scanf("%s", s + 1);
        for(int i = 1; i <= n; i++){
            int t = s[i] - '0';
            presum[i] = presum[i - 1] + t;
        }

        bool flag = false;
        if(presum[n] == 0){
            printf("YES\n");
            continue;
        }
        //cout<<presum[n]<<endl;
        for(int i = 1; i <= n; i++){
            if(presum[i] == 0){
                continue;
            }
            if(presum[n] % presum[i]){
                continue;
            }
            else{
                int k = presum[n] / presum[i];
                if(k == 2){
                    flag = true;
                    break;
                }
                int t = 2;
                for(int j = i + 1; j <= n; j++){
                    if(presum[j] == (t) * presum[i]){
                        //cout<<k<<endl;
                        //cout<<i<<" "<<presum[i]<<endl;
                        //cout<<j<<" "<<presum[j]<<" "<<endl;
                        t++;
                    }
                    if(k == t){
                        flag = true;
                        break;
                    }
                }
                if(flag){
                    break;
                }
            }
        }

        if(flag){
            printf("YES\n");
        }
        else{
            printf("NO\n");
        }
    }
    return 0;
}

 

 

D-Vasya and Triangle

三角形公式S=(1/2)*(x1y2*1+x2y3*1+x3y1*1-x1y3*1-x2y1*1-x3y2*1) =1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

当2*n*m/k不是整数时一定没有解

有一个点一定是原点 否则一定可以将这个三角形平移到原点

剩下两个点一个点的横坐标和另一个点的纵坐标一定是0 因为这样就已经足够取尽n*m/2中的所有整数了

接下来就是如何凑出x2和y3了 使得x2 * y3 = 2 * n * m / k

应该想到的是gcd 

假设a = gcd(2 * n, k) 那么S = (2 * n / a) * m / (k / a)

将他们分配一下 x2 = (2 * n / a), y3 = (m * a / k)

如果x2 或 y3超出范围限制了 就考虑把2这个系数挪一下就好了

#include <bits/stdc++.h>
using namespace std;
typedef long long int LL;

LL n, m, k;

LL gcd(LL a, LL b)
{
    if(b == 0)return a;
    return gcd(b, a % b);
}

int main()
{
    while(scanf("%lld%lld%lld", &n, &m, &k) != EOF){
        if((2 * n * m) % k){
            printf("NO\n");
        }
        else{
            LL x1, x2, x3, y1, y2, y3;
            x1 = 0ll; y1 = 0ll; x3 = 0ll; y2 = 0ll;
            LL a = gcd(2 * n, k);
            //cout<<a<<endl;
            y3 = a * m / k;
            x2 = 2 * n / a;
            //cout<<x2<<" "<<y3<<endl;
            if(x2 > n || y3 > m){
                y3 *= 2;
                x2 /= 2;
            }
            /*if(n % k == 0){
                y3 = m;
                x2 = 2 * n / k;
            }
            else if(m % k == 0){
                x2 = n;
                y3 = 2 * m / k;
            }
            else{
                /*int a = n / gcd(n, k);
                int b = m / gcd(m, k);
                if(a * 2 < n){
                    x2 = a * 2;
                    y3 = b;
                }
                else if(a * 2 < m){
                    y3 = a * 2;
                    x2 = b;
                }
                else if(b * 2 < n){
                    x2 = b * 2;
                    y3 = a;
                }
                else if(b * 2 < m){
                    y3 = b * 2;
                    x2 = a;
                }
            }*/
            printf("YES\n");
            printf("%lld %lld\n", x1, y1);
            printf("%lld %lld\n", x2, y2);
            printf("%lld %lld\n", x3, y3);
        }
    }

    return 0;
}