990——Educational Codeforces Round 45 (Rated for Div. 2)

http://codeforces.com/contest/990

codeforces

考试完成
A：模拟
B：模拟
C：模拟计数，统计左括号多x的个数，右括号同理，当前左括号多x，+cntr[x]

D：特判
E：调和级数，枚举k即可，验证是否可行（复杂度nlojn，类似于nlogn求素数）
F：
G:

代码T4：

    int n = read(),a = read(),b = read();
    if (n==1) {
        if (a==1&&b==1) printf("YES\n0");
        else printf("NO");
        return 0;
    }
    if (n==2) {
        if (a==1 && b==2) { printf("YES\n01\n10"); }
        else if (a==2 && b==1) printf("YES\n00\n00");
        else printf("NO");
        return 0;
    }
    if (n==3) {
        if (a==1 && b==3) printf("YES\n011\n101\n110");
        else if (a==1 && b==2) printf("YES\n001\n001\n110");
        else if (a==2 && b==1) printf("YES\n001\n000\n100");
        else if (a==3 && b==1) printf("YES\n000\n000\n000");
        else printf("NO");
        return 0;
    }
    if (b >= 2 && a!=1) {printf("NO");return 0;}
    if (a==1) {
        printf("YES\n");
        for (int i=1; i<=(n-b); ++i) mp[i][i+1] = mp[i+1][i] = 1;
        for (int i=1; i<=n; ++i) {
            for (int j=1; j<=n; ++j) 
                if (i!=j)printf("%d",!mp[i][j]);
                else printf("0");
            printf("\n");
        }
    }
    if (a >= 2) {
        if (b != 1) printf("NO");
        else {
            printf("YES\n");
            for (int i=1; i<=(n-a); ++i) mp[i][i+1] = mp[i+1][i] = 1;
            for (int i=1; i<=n; ++i) {
                for (int j=1; j<=n; ++j) 
                    printf("%d",mp[i][j]);
                printf("\n");
            }
        }
    }