[英]C++ Nested For Loops
我認為這會比最初計划的要容易。 我正在嘗試使此圖像脫離嵌套的for循環:
任何建議或解決方案都將有所幫助。
#include <iostream>
using namespace std;
int main()
{
for(int i=0; i<1;i++)
{
cout<<i+1<<endl;
for(int j=0;j<2;j++)
{
cout<<j+1;
}
}
cout<<"\n";
for(int k=0; k<3; k++)
{
cout<<k+1;
}
cout<<"\n";
for(int l=0; l<4; l++)
{
cout<<l+1;
}
cout<<"\n";
for(int m=4; m>0; m--)
{
cout<<m;
}
cout<<"\n";
for(int n=3; n>0; n--)
{
cout<<n;
}
cout<<"\n";
for(int o=2; o>0; o--)
{
cout<<o;
}
cout<<"\n";
for(int p=0; p<1; p++)
{
cout<<p+1;
}
cin.get();
return 0;
}
這是為您提供的C解決方案=)
#include <stdio.h>
#include <string.h>
int main(void) {
char forward[5] = "1";
char reverse[5] = "4321";
int i;
for( i = 1; i <= 4; i++ ) {
printf( "%s\n", forward );
forward[i] = forward[i-1]+1;
}
for( i = 0; i < 4; i++ ) printf( "% 4s\n", reverse+i );
}
如果您想要一個非常緊湊且不太實用的解決方案,我決定嘗試一下。
int length = 4;
for(int i = 0; i < 2; i++)
{
for(int j = 0; j < length; j++)
{
for(int k = 0; k < (i == 1 ? length - j : j + 1); k++)
{
if(i == 1 && k == 0)
for(int x = 0; x < j; x++)
cout << " ";
cout << (i == 1 ? (length - k) - j : k + 1);
}
cout << endl;
}
}
其中length
是從1
到length
的迭代次數。
好的,因為paddy發布了他的C解決方案(有人說“緊湊” = P)...
#include <stdio.h>
int main()
{
char line[] = "1234321";
int i=0;
for (; i<4; printf("%.*s\n",++i,line));
for (i=0;i<4; printf("%4s\n",line+3+i++));
return 0;
}
輸出量
1
12
123
1234
4321
321
21
1
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
/*first half: From level1 = 1 to less than 5, level2 = 1 to less than or equal to level1, print level2 */
for(int level1 = 1; level1 < 5; level1++) {
for(int level2 = 1; level2 <= level1; level2++){
cout << level2;
}
cout << '\n'; /*prints the newline AFTER each iteration of the `level2` loop*/
}
/*second half: reverse the logic of part1, but also add spaces in the beginning */
for(int level1 = 4; level1 > 0; level1--){
for(int interim = 4; interim > level1; interim--) cout << ' ';
for(int level2 = level1; level2 > 0; level2--){
cout << level2;
}
cout << '\n';
}
return 0;
}
如果您不了解此代碼,請告訴我:-)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.