Idea: Preprocess all the numbers to be used and put them into a large string,
and then deal with the coordinate changes according to the situation.
code show as below:
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#define ll long long
#define inf 0x3f3f3f3f
#define sd(a) scanf("%d",&a)
#define sdd(a,b) scanf("%d%d",&a,&b)
#define cl(a,b) memset(a,b,sizeof(a))
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define sddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define dbg() printf("aaa\n")
using namespace std;
const int maxn=1010;
char mp[maxn][maxn];
char c[maxn*10];
int main() {
//初始化
int p=0;
for(int i=1;i<=999;i++){
if(i<=9){
c[++p]='0'+i;
}else if(i<=99){
c[++p]='0'+i/10;
c[++p]='0'+i%10;
}else if(i<=999){
c[++p]='0'+i/100;
c[++p]='0'+i/10%10;
c[++p]='0'+i%10;
}
}
int n;
sd(n);
rep(i,1,n){
rep(j,1,n+i-1){
mp[i][j]='.';
}
}
int k=0;
int i=1,j=n;
while(1){
mp[i][j]=c[++k];
if(i<n&&j<=n){
i++,j--;
}else if(i==n&&j!=2*n-1){
j++;
}else{
i--,j--;
}
if(i==1&&j==n) break;
}
rep(i,1,n){
rep(j,1,n+i-1){
putchar(mp[i][j]);
}putchar('\n');
}
return 0;
}
This problem requires you to output an isosceles triangle composed of numbers to the console.
The specific steps are:
-
First use the natural numbers of 1, 2, 3, ... to make a long enough string
-
Fill the three sides of the triangle with this string. Start at the top vertex and fill in counter-clockwise.
For example, when the triangle height is 8:1
2 1
3 8
4 1
5 7
6 1
7 6
891011121314151
Input
A positive integer n (3<n<300) representing the height of the triangle
Output
An isosceles triangle filled with numbers.
For the convenience of evaluation, we require spaces to be replaced with ".".
sample input
5
Sample output
…1
…2.1
…3…2
.4…1
567891011