Educational Codeforces Round 20 A

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就是一个n*n方阵，有k个1放在里面，然后要求对角线对称且字典序最大，如果k>n*n就输出-1，否则就看半边，遍历一遍，k足够放两个1的话，就放一个1，然后每次换行的时候要判断还剩几个1没放，剩一个就放在对角线，如果还剩两个的话就在对角线连放两个1，然后再把另外半边对称放上就ok了（详见代码）。

#include<iostream>#include<cstdio>#include<cstring>#include<ctime>#include<algorithm>#include<cstdlib>#include<cmath>#include<set>#include<bitset>#include<map>#include<stack>#include<queue>#include<vector>#include<utility>#define INF 0x3f3f3f3f#define inf 2*0x3f3f3f3f#define llinf 1000000000000000000#define pi acos(-1.0)#define mod 1000000007#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lrt rt<<1#define rrt rt<<1|1#define rep(i,a,b) for(int i=(a);i<(b);i++)#define per(i,a,b) for(int i=(b)-1;i>=(a);i--)#define mem(a,b) memset(a,b,sizeof(a))#define lb(x) (x&-x)#define gi(x) scanf("%d",&x)#define gi2(x,y) scanf("%d%d",&x,&y)#define gll(x) scanf("%lld",&x)#define gll2(x,y) scanf("%lld%lld",&x,&y)#define gc(x) scanf("%c",&x)using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int,int>P;/***********************************************/int n,k,mm[105][105],ss=0;int main(){    gi2(n,k);    if(k>n*n)return cout<<-1,0;    rep(i,0,n)    {        if(k<=2*ss)break;        mm[i][i]=1;k--;        if(k-ss*2==1){mm[i+1][i+1]=1;break;}        rep(j,i+1,n)        {            if(k-ss*2>=2)mm[i][j]=1,ss++;        }    }    rep(i,0,n)    {        rep(j,i+1,n)        {            mm[j][i]=mm[i][j];        }    }    rep(i,0,n)    {        rep(j,0,n)        {            printf("%d ",mm[i][j]);        }        printf("\n");    }    return 0;}

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