11 solutions

  • 0
    @ 2025-2-2 19:59:00

    结构体够使呆码

    #include<bits/stdc++.h>
using namespace std;
struct fba{
   unsigned short inxi,inxj,val,f;
}lis[6][6],maxlis[6],maxli[6]; 
int main(){
   for(short i=1;i<=5;i++){
      maxli[i].val=SHRT_MAX*2-1;
      for(short j=1;j<=5;j++){
         cin>>lis[i][j].val;
         lis[i][j].inxi=i;
         lis[i][j].inxj=j;
         if(lis[i][j].val>maxlis[i].val)
           {maxlis[i].val=lis[i][j].val;
            maxlis[i].inxi=i;maxlis[i].inxj=j;}
         }
      lis[maxlis[i].inxi][maxlis[i].inxj].f++;
   }
   for(int j=1;j<=5;j++){
      for(int i=1;i<=5;i++){
         if(lis[i][j].val<maxli[j].val){
            maxli[j].val=lis[i][j].val;
            maxli[j].inxi=i;maxli[j].inxj=j;}
      }
      lis[maxli[j].inxi][maxli[j].inxj].f++;
   }
   for(int i=1;i<=5;i++){
      for(int j=1;j<=5;j++){
         if(lis[i][j].f==2) 
         {cout<<i<<' '<<j<<' '<<lis[i][j].val;
         return 0;}
      }
   }
   cout<<"not found";
   return 0;}
    • 0
      @ 2024-11-12 19:51:40

      屎山代码

      #include<bits/stdc++.h>
using namespace std;
int ppp[110][110];
int main(){
	for(int x=0;x<5;x++)for(int y=0;y<5;y++)cin>>ppp[x][y];
	for(int x=0;x<5;x++){
		int max=-100000;
		for(int y=0;y<5;y++)if(max<ppp[x][y])max=ppp[x][y];
		for(int y=0;y<5;y++)if(max==ppp[x][y]){
			int min=100000;
			for(int xx=0;xx<5;xx++)if(min>ppp[xx][y])min=ppp[xx][y];
			if(max==min){
				cout<<x+1<<" "<<y+1<<" "<<max;
				return 0;
			}
		}
	}
	cout<<"not found";
	return 0;
}

      N3+2*N2

      • 0
        @ 2024-10-23 16:16:32 
        #include<bits/stdc++.h> 

using namespace std;
int a[1001][1001];
int main(){
	for(int i=1;i<=5;i++){
		for(int j=1;j<=5;j++){
			cin>>a[i][j];
		}
	}
	for(int i=1;i<=5;i++){
		int maxs=0,mins=1e9+7;//每行最大的，每列最小的 
		int i2=i,j2;//j2存每行最大的下标 
		//遍历每行 
		for(int j=1;j<=5;j++){
			if(a[i][j]>maxs){
				maxs=a[i][j];
				j2=j;
			}
		}
		//遍历最大值所在的列，寻找最小 
		for(int j=1;j<=5;j++){
			mins=min(mins,a[j][j2]);
		}
		if(mins==maxs){//如果最大==最小，直接输出结束 
			cout<<i2<<" "<<j2<<" "<<a[i2][j2];
			return 0;
		}
	
	}
	cout<<"not found";//实在没有就输出“not found” 
	return 0;
}
        • 0
          @ 2024-8-24 16:11:44 
          #include<bits/stdc++.h>
using namespace std;
int main(){
	int a[10][10];
	for(int i=1;i<=5;i++){
		for(int j=1;j<=5;j++){
			cin>>a[i][j];
		}
	}
	for(int i=1;i<=5;i++){
		for(int j=1;j<=5;j++){
			if(a[i][j]>=a[i][1] && a[i][j]>=a[i][2] && a[i][j]>=a[i][3] && a[i][j]>=a[i][4] && a[i][j]>=a[i][5]){
				if(a[i][j]<=a[1][j] &&  a[i][j]<=a[2][j] &&  a[i][j]<=a[3][j] &&  a[i][j]<=a[4][j] &&  a[i][j]<=a[5][j]){
					cout<<i<<' '<<j<<' '<<a[i][j];
					return 0;
				}
			}
		}
	}
	cout<<"not found";
	return 0; 
}
          • 0
            @ 2024-2-4 13:17:26 
            #include<bits/stdc++.h>
using namespace std;
int a[10][10],h[10],l[10],maxh,minl;
bool flag = true;

int main () {
	for (int i = 1; i <= 5; i++){
		for (int j = 1; j <= 5; j++) {
			cin >> a[i][j];
		}
	}	
	for (int i = 1; i <= 5; i++) {
		maxh = INT_MIN;
		for (int j = 1; j <= 5; j++){
			if (a[i][j]>maxh) {
				h[i] = j;
				maxh = a[i][j];
			}
		}
	}
	for (int i = 1; i <= 5; i++) {
		minl = INT_MAX;
		for (int j = 1; j <= 5; j++)
			if (a[j][i] < minl) {
				l[i] = j;
				minl = a[j][i];
			}
	}
	for (int i = 1; i <= 5; i++)
		if (i==l[h[i]]) {
			cout << i << " " << h[i] << " " << a[i][h[i]];
			flag = false;
		}
	if (flag) {
		cout << "not found" << endl;
	}
	return 0;
}
            • 0
              @ 2024-1-28 10:40:28 
              #include<bits/stdc++.h>
using namespace std;
int a[5][5];
int x,y;
int ma=0,mi=10000000;
int main(){
	for(int i=0;i<5;i++) for(int j=0;j<5;j++) cin>>a[i][j];
	for(int i=0;i<5;i++){
		int j,k;
		ma=0,mi=10000000;
		for(j=0;j<5;j++) ma=max(ma,a[i][j]);
		for(j=0;j<5;j++) if(a[i][j]==ma) for(k=0;k<5;k++) if(a[k][j]<mi) mi=a[k][j],x=k,y=j;	
		if(mi==ma){
			cout<<x+1<<' ';
			cout<<y+1<<' ';
			cout<<ma<<"\n";
			break;
		}
		if(i==4&&mi!=ma) cout<<"not found";
	}
	return 0;
}

              极致压行

              • -1
                @ 2023-12-24 18:13:18

                一个完美的做法

                #include<bits/stdc++.h>
using namespace std;
int a[10][10],h[10],l[10],maxh,minl;
bool pd=true;
int main(){
    for(int i=1;i<=5;i++)
        for(int j=1;j<=5;j++) cin>>a[i][j];
    for(int i=1;i<=5;i++){
        maxh=-20000000; 
        for(int j=1;j<=5;j++) if(a[i][j]>maxh) h[i]=j,maxh=a[i][j];
    }
    for(int i=1;i<=5;i++)
    {
        minl=20000000;
        for(int j=1;j<=5;j++) if(a[j][i]<minl) l[i]=j,minl=a[j][i];
    }
    for(int i=1;i<=5;i++)
        if(i==l[h[i]]){
            cout<<i<<" "<<h[i]<<" "<<a[i][h[i]];
            pd=false;
        }
    if(pd) cout<<"not found"<<endl;
    return 0;
 }
                • -2
                  @ 2024-2-28 13:06:50 
                  using namespace std;
int a[10][10],h[10],l[10],maxh,minl;
bool pd=true;
int main()
{
for(int i=1;i<=5;i++)
for(int j=1;j<=5;j++) cin>>a[i][j];
for(int i=1;i<=5;i++)
{
maxh=-1;
for(int j=1;j<=5;j++)
if(a[i][j]>maxh)
{
h[i]=j;
maxh=a[i][j];
}
}
for(int i=1;i<=5;i++)
{
minl=99999;
for(int j=1;j<=5;j++)
if(a[j][i]<minl)
{
l[i]=j;
minl=a[j][i];
}
}
for(int i=1;i<=5;i++)
if(i==l[h[i]])
{
cout<<i<<" "<<h[i]<<" "<<a[i][h[i]];
pd=false;
}
if(pd) cout<<"not found"<<endl;
return 0;
}
                  • -2
                    @ 2023-10-4 15:20:52

                    #include<bits/stdc++.h> using namespace std; int main(){ int a[5][5]; for(int i=0;i<5;i++){ for(int j=0;j<5;j++){ cin>>a[i][j]; } }for(int i=0;i<5;i++){ for(int j=0;j<5;j++){ int b=0; int c=0; for(int k=0;k<5;k++){ if(a[i][k]>a[i][j]){ b++; } } for(int k=0;k<5;k++){ if(a[k][j]<a[i][j]){ c++; } } if(b0&&c0){ cout<<i+1<<" "<<j+1<<" "<<a[i][j]; return 0; } } }

                    }

                    • -2
                      @ 2023-10-4 14:53:42 
                      #include <bits/stdc++.h>
using namespace std;
int main(int argc, char **argv){
	int x,y,a[5][5];
	for (int i = 0;i < 5;i++){
		for (int j = 0;j < 5;j++){
			cin >> a[i][j];
		}
	}
	for (int i = 0;i < 5;i++){	// 遍历行
		int mxh = -1;bool f = 1;
		for (int j = 0;j < 5;j++){	// 遍历列
			if (mxh < a[i][j]){
				x = i + 1;
				y = j + 1;
				mxh = a[i][j];
			}
		}
		for (int j = 0;j < 5;j++){
			if (mxh > a[j][y - 1]){
				f = 0;
			}
		}
		if (f){
			printf("%d %d %d",x,y,mxh);
			return 0;
		}
	}
	printf("not found");
	return 0;
}
                      • -2
                        @ 2023-9-29 11:57:58

                        `

                        #include<bits/stdc++.h>
using namespace std;
int a[10][10],h[10],l[10],maxh,minl;
//a数组存矩阵，h、l含义见上文，maxh代表行的最大值，minl代表列的最小值 
bool pd=true;
int main()
{
    for(int i=1;i<=5;i++)
        for(int j=1;j<=5;j++) cin>>a[i][j];
    for(int i=1;i<=5;i++)
    {
        maxh=-1;//重置最大值为-1 
        for(int j=1;j<=5;j++)
            if(a[i][j]>maxh)
            {
                h[i]=j;//找到最大值所在的列 
                maxh=a[i][j];
            }
    }
    for(int i=1;i<=5;i++)
    {
        minl=99999;//重置最小值为99999 
        for(int j=1;j<=5;j++)
            if(a[j][i]<minl)
            {
                l[i]=j;//找到最小值所在的行 
                minl=a[j][i];
            }
    }
    for(int i=1;i<=5;i++)
        if(i==l[h[i]])
        //对比是否有同时满足 行最大，列最小 
        {
            cout<<i<<" "<<h[i]<<" "<<a[i][h[i]];
            pd=false;
        }
    if(pd) cout<<"not found"<<endl;
    //若没有，则输出 "not found"
    return 0;
 }
                        • 1

                        计算鞍点

                        Information

                        ID
                        608
                        Time
                        1000ms
                        Memory
                        256MiB
                        Difficulty
                        5
                        Tags
                        # Submissions
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                        Accepted
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