codeforces 984补题

codeforces 984a

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n;
 4 int num[1010];
 5 int main(){
 6     scanf("%d",&n);
 7     for(int i=1;i<=n;i++) scanf("%d",&num[i]);
 8     sort(num+1,num+n+1);
 9     if(n%2==1) printf("%d",num[(n+1)/2]);
10     else printf("%d",num[n/2]);
11     return 0;
12 }

codeforces 984b

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n,m;
 4 char sz[110][110];
 5 int divs[8][2]={{0,1},{0,-1},{1,0},{-1,0},{-1,-1},{1,-1},{-1,1},{1,1}};
 6 int main(){
 7     while(scanf("%d%d",&n,&m)!=EOF){
 8         for(int i=1;i<=n;i++) scanf("%s",sz[i]+1);
 9         for(int i=1;i<=n;i++){
10             for(int j=1;j<=m;j++){
11                 if(sz[i][j]==‘*‘) continue;
12                 else if(sz[i][j]==‘.‘){
13                     for(int k=0;k<8;k++){
14                         int x=i+divs[k][0],y=j+divs[k][1];
15                         if(x>=1&&x<=n&&y>=1&&y<=m){
16                             if(sz[x][y]==‘*‘){printf("NO\n");return 0;}
17                         }
18                     }
19                 }else{
20                     int num=sz[i][j]-‘0‘;
21                     for(int k=0;k<8;k++){
22                         int x=i+divs[k][0],y=j+divs[k][1];
23                         if(x>=1&&x<=n&&y>=1&&y<=m){
24                             if(sz[x][y]==‘*‘){num--;}
25                         }
26                     }
27                     if(num!=0) { printf("NO\n");return 0;}
28                 }
29                 //if(i==n&&j==m) printf("sbsbsbsb\n");
30             }
31         }
32         //printf("sbbsbsbsbsb\n");
33         printf("YES\n");
34     }
35     return 0;
36 }

先跳过c题补一下d题

codeforces984d 先贴一下链接:看了别人的代码: https://blog.csdn.net/qq_39599067/article/details/80335949

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n,q;
 4 int num[5010];
 5 int dp[5010][5010],f[5010][5010];
 6 int main(){
 7     scanf("%d",&n);
 8     for(int i=1;i<=n;i++){
 9         scanf("%d",&num[i]);
10         dp[i][i]=f[i][i]=num[i];
11     }
12     for(int j=2;j<=n;j++){
13         for(int i=j-1;i>=1;i--){
14             f[i][j]=f[i][j-1]^f[i+1][j];
15             dp[i][j]=max(f[i][j],max(dp[i][j-1],dp[i+1][j]));
16         }
17     }
18     scanf("%d",&q);
19     while(q--){
20         int l,r;
21         scanf("%d%d",&l,&r);
22         cout<<dp[l][r]<<endl;
23     }
24     return 0;
25 }

codeforces c题 还是贴一个别人的链接  https://blog.csdn.net/xs18952904/article/details/80331434

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 int n;
 5 LL p,q,c;
 6
 7 LL gcd(LL a,LL b){
 8     if(b==0) return a;
 9     else return gcd(b,a%b);
10 }
11
12 int main(){
13     scanf("%d",&n);
14     while(n--){
15         scanf("%lld%lld%lld",&p,&q,&c);
16         //printf("%d\n",gcd(p,q));
17         q=q/gcd(p,q);
18         //printf("%d\n",q);
19         while(gcd(q,c)!=1){
20             LL tmp=gcd(q,c);
21             while(q%tmp==0) q=q/tmp;
22         }
23         if(q==1) printf("Finite\n");
24         else printf("Infinite\n");
25     }
26     return 0;
27 }

原文地址：https://www.cnblogs.com/pandaking/p/9739624.html