暴力枚举,但简单易懂

刹那(。・∀・)ノ゙  •  2年前


反反复复找了n次bug 最后采用了暴力枚举,虽然笨,但胜在简单易懂 if战神(bushi

include <stdio.h>

include <string.h>

int ans[101][101]={0}; char a[101][101]; int main() {

int m, n;
scanf("%d%d", &n, &m);
for (int  i = 1; i <= n; i++) 
{
	for (int  k = 1; k <= m+1; k++)
	{
		scanf("%c", &a[i][k]);		
	}
}
for (int i = 1; i <= n; i++) 
{
	for (int k = 1; k <= m+1; k++) 
	{
		
			if (a[i + 1][k] == '*')
			{
				ans[i][k]++;
			}
			if(a[i - 1][k] == '*' )
			{
				ans[i][k]++;
			}
			if(a[i][k - 1] == '*')
			{
				ans[i][k]++;
			}
			if(a[i+1][k + 1] == '*')
			{
				ans[i][k]++;
			}
			if(a[i+1][k-1]=='*')
			{
				ans[i][k]++;
			}
			if(a[i-1][k+1]=='*')
			{
				ans[i][k]++;
			}
			if(a[i-1][k-1]=='*')
			{
				ans[i][k]++;
			}
			if(a[i][k+1]=='*')
			{
				ans[i][k]++;
			}
		
	}	
}
for(int i=1;i<=n;i++)
{
	for(int k=2;k<=m+1;k++)
	{
		if(a[i][k]=='*')
		{
			printf("*");
		}
		else
		{
			printf("%d",ans[i][k]);
		}
	}
	printf("\n");
}
return 0;

}


评论:

include<stdio.h> int main() {

int n,s,p,a,b,i,j,k; s=0; p=0; scanf("%d",&n); for(i=3;i<n;i++){ a=i; b=n-a; s=0; p=0; for(j=2;j<a;j++){ if(a%j==0){ s++; } } for(k=2;k<b;k++){ if(b%k==0){ p++; } } if(s==0&&p==0){ break; } } printf("%d ",n); printf("= %d + %d",a,b); return 0; }


root  •  2年前

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