Educational CF 84（Div2）

A

题意：n是否能被k个不同的正奇数表示。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 50000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a < b ? -1 : 1;
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        LL n, k;
        scanf("%lld%lld", &n, &k);
        if(n >= k * k && (n + k) % 2 == 0) printf("YES
");
        else printf("NO
");
    }
    return 0;
}

B

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 100000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a < b ? -1 : 1;
}
int g[MAXN], b[MAXN];
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i){
            g[i] = b[i] = 0;
        }
        for(int i = 1; i <= n; ++i){
            int k, x;
            scanf("%d", &k);
            while(k--){
                scanf("%d", &x);
                if(!g[i] && !b[x]){
                    g[i] = x;
                    b[x] = i;
                }
            }
        }
        int ans1 = 0;
        int ans2 = 0;
        for(int i = 1; i <= n; ++i){
            if(!g[i]){
                ans1 = i;
                break;
            }
        }
        for(int i = 1; i <= n; ++i){
            if(!b[i]){
                ans2 = i;
                break;
            }
        }
        if(ans1){
            printf("IMPROVE
");
            printf("%d %d
", ans1, ans2);
        }
        else{
            printf("OPTIMAL
");
        }
    }
    return 0;
}

C

题意：n*m的格子，有k个点，要求每个点都必须经过某个点。每次向某个方向移动，所有点是一起移动的，如果到了边界，则原地不动。问是否能在2mn步内使每个点都经过某个指定的点。

分析：将k个点都移到左上角，再遍历所有格子。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 50000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a < b ? -1 : 1;
}
int main(){
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    int x, y;
    for(int i = 0; i < 2 * k; ++i){
        scanf("%d%d", &x, &y);
    }
    printf("%d
", (n - 1) + (m - 1) + (m - 1) * n + (n - 1));
    for(int i = 0; i < n - 1; ++i) printf("U");
    for(int i = 0; i < m - 1; ++i) printf("L");
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m - 1; ++j){
            if(i % 2 == 0) printf("R");
            else printf("L");
        }
        if(i != n - 1) printf("D");
    }
    printf("
");
    return 0;
}