#include<stdio.h>
int main(){
int h;
scanf("%d",&h);
for(int i=1;i<=h;i++){
for(int j=1;j<=3*h-2;j++){
if(j <= 2*(h-i))
printf(" ");
else
printf("*");
}
printf("\n");
}
return 0;
}
注意，scanf在连续输入的时候不能带空格
#include<stdio.h>
int main(){
int n;
char a[80][80];
char A,B;
scanf("%d%c%c",&n,&B,&A);
/*
scanf("%d",&n);
getchar();
scanf("%c",&B);
getchar();
scanf("%c",&A);
*/
a[0][0]=a[0][n-1]=' ';// reset the four corner
a[n-1][0]=a[n-1][n-1]=' ';
int even = ((n-1)/2)%2;//1->A 0->B
char edge=A;
if (even == 0){
edge=B;
}
else{
edge=A;
}
//the outest level
for(int i=1;i<=n-2;i++){
a[0][i]=edge;
a[n-1][i]=edge;
a[i][0]=edge;
a[i][n-1]=edge;
}
for(int c=2;c<=(n+1)/2;c++){
if(edge == A){
edge = B;
}
else{
edge = A;
}
for(int i=c-1;i<=n-c;i++){
a[c-1][i]=edge;
a[n-c][i]=edge;
a[i][c-1]=edge;
a[i][n-c]=edge;
}
}
for(int i=0;i<n;i++){
for(int j =0;j<n;j++){
printf("%c",a[i][j]);
}
printf("\n");
}
return 0;
}
牛客网1161：
题目：
The input contains multiple test cases.
The first line of each case is an integer N, representing the size of the template is NN (N could only be 3, 4 or 5).
Next N lines describe the template.
The following line contains an integer Q, which is the Scale Level of the picture.
Input is ended with a case of N=0.
It is guaranteed that the size of one picture will not exceed 30003000.
输入包含多个测试用例。
每种情况的第一行都是一个整数N，表示模板的大小为NN (N只能是3、4或5)。
接下来的N行描述模板。
下一行包含一个整数Q，它是图片的缩放级别。
输入以N=0的情况结束。
保证一张图片的尺寸不超过30003000。
输入：
3
# #
#
# #
1
3
# #
#
# #
3
4
OO
O O
O O
OO
2
0
输出：
# #
#
# #
# # # # # # # #
# # # #
# # # # # # # #
# # # #
# #
# # # #
# # # # # # # #
# # # #
# # # # # # # #
# # # #
# #
# # # #
# #
#
# #
# # # #
# #
# # # #
# # # # # # # #
# # # #
# # # # # # # #
# # # #
# #
# # # #
# # # # # # # #
# # # #
# # # # # # # #
OO OO
O OO O
O OO O
OO OO
OO OO
O O O O
O O O O
OO OO
OO OO
O O O O
O O O O
OO OO
OO OO
O OO O
O OO O
OO OO
习题2.5 hello world for u
#include <stdio.h>
#include <string.h>
#define N 27
int main(void)
{
char str[81];
char out[N][N];
int n1,n2,n3;
while(scanf("%s",str) != EOF){
int len=strlen(str);
int a = len/3;
int b = len%3;
if(b!=0){
n1 = a;
n2 = a + b;
}
else{
n1 = a-1;
n2 = a+2;
}
for(int i=0;i<N;i++)
for(int j=0;j<N;j++){
out[i][j]=' ';
}
for(int i =0; i<n1;i++)
out[i][0]=str[i];
for(int i = 0; i<n2; i++)
out[n1][i]=str[i+n1];
for(int i = 0;i < n1;i++)
out[n1-1-i][n2-1]=str[n1+n2+i];
for(int i=0;i<=n1;i++){
for(int j=0;j<n2;j++){
printf("%c",out[i][j]);
}
printf("\n");
}
}
return 0;
}
习题2.6 今年的第几天
注意 闰年判断和月份判断
闰年：能被4整除但不能被100整除 or 能被400整除
#include <stdio.h>
#include <string.h>
int mon[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int runYear(int a){
if((!(a%4) && (a%100)) || !(a%400))
return 1;
return 0;
}
int main(void)
{
int year,month,day,sum;
sum=0;
while(scanf("%d%d%d", &year,&month,&day) != EOF)
{ getchar();
for(int i=1;i<month;i++){
sum += mon[i];
}
if(runYear(year) && month>2)
sum += day+1;
else
sum += day;
printf("%d\n",sum);
sum=0;
}
return 0;
}
例题2.7 打印日期
注意输出格式：个位数时1显示为01, 5显示为01， printf("%02d", month);
#include <stdio.h>
#include <string.h>
int nis[13]={0,31,59,90,120,151,181,212,243,273,304,334,365};
int is[13]={0,31,60,91,121,152,182,213,244,274,305,335,366};
int runYear(int a){
if((!(a%4) && (a%100)) || !(a%400))
return 1;
return 0;
}
int main(void)
{
int n,year,month,day,sum;
sum=0;
while(scanf("%d%d", &year,&n) != EOF)
{ getchar();
if(runYear(year))
for(int i=1;i<=12;i++){
if(n <=is[i]){
month = i;
day = n-is[i-1];
break;
}
else if(n > is[i] && n<is[i+1]){
month = i+1;
day = n-is[i];
break;
}
else
continue;
}
else{
for(int i=1;i<=12;i++){
if(n <=nis[i]){
month = i;
day = n-nis[i-1];
break;
}
else if(n > nis[i] && n<nis[i+1]){
month = i+1;
day = n-nis[i];
break;
}
else
continue;
}
}
printf("%04d-%02d-%02d\n",year,month,day);
}
return 0;
}
这个代码更简洁
#include <stdio.h>
#include <string.h>
int daytab[2][13]={
{0,31,28,31,30,31,30,31,31,30,31,30,31},
{0,31,29,31,30,31,30,31,31,30,31,30,31}
};
int runYear(int a){
if((!(a%4) && (a%100)) || !(a%400))
return 1;
return 0;
}
int main(void)
{
int n,year,month,day;
while(scanf("%d%d", &year,&n) != EOF)
{ getchar();
month = 0;
int row = runYear(year);
//确定月
while(n > daytab[row][month]){
n -= daytab[row][month];
month++;
}
//确定日
day = n;
printf("%04d-%02d-%02d\n",year,month,day);
}
return 0;
}