Level 7: Yang Hui Triangle
The code of the Yang Hui triangle refers to this blogger's article:
Title description: Do you still remember the Yang Hui triangle that you learned in middle school? The specific definition will not be described here, you can refer to the following graphics:
output
Print out 10 lines of the Yang Hui triangle figure. See the topic description section for the format. Each integer is followed by a space to separate the integers
#include <stdio.h>
int main(void)
{
int data[10][10];//define a two-dimensional array
for(int i=0;i<10;i++){
for(int j=0;j<10;j++){
data[i][j]=1;//The two-dimensional array is all initialized to 1
}
}
//Process the data in the middle: starting from the 3rd row and the 2nd column, the current number = the number in the previous row + the number in front of the previous row
for(int k=1;k<10;k++){
for(int p=1;p<k;p++){
data[k][p]=data[k-1][p]+data[k-1][p-1];
}
}
//Output Data
for(int i=0;i<10;i++){
for(int j=0;j<=i;j++){
printf("%d ",data[i][j]);
}
printf("\n");
}
return 0;
}
Level 6: Delete the maximum value
Title description: Input 10 different integers and save them in an array, find the largest element and delete it, and output the deleted array
#include<stdio.h>
int main(void)
{
/*********Begin*********/
int a[10];
for(int i=0;i<10;i++){
scanf("%d",&a[i]);
}
int max=a[0],num=0;
for(int i=0;i<10;i++){//find the maximum value max, num
if(a[i]>max){
max=a[i];
num=i;
}
}
for(int i=num;i<9;i++){//remove the largest value a[num]
a[i]=a[i+1];
}
//output array
for(int i=0;i<9;i++){
printf("%d ",a[i]);
}
/*********End**********/
return 0;
}
Level 5: Saddle Point
Title description: Find the "saddle point" of a two-dimensional array Array with m rows and n columns, that is, the element at this position is the largest on the row and the smallest on the column, where 1<=m, n<=10.
enter
The input data has multiple lines, the first line has two numbers m and n, and there are m lines below, each line has n numbers.
output
Output the saddle point in the following format: Array[i][j]=x where x represents the saddle point, i and j are the array row and column subscripts where the saddle point is located, and we stipulate that the array subscript starts from 0. A two-dimensional array does not necessarily have a saddle point, please output None at this time and we guarantee that there will not be two saddle points, for example:
#include<stdio.h>
int main(void)
{
/*********Begin*********/
int m,n;
scanf("%d %d",&m,&n);
int a[m][n];
//enter
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
scanf("%d",&a[i][j]);
}
}
int MAX=a[0][0],row=0,col=0;
int MIN;
//Determine whether it is a saddle point
//Determine whether the largest number in the same row is the smallest number in the same column
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
//find the largest number in a row
if(a[i][j]>MAX){
MAX= a[i][j];
//printf("%d***\n",MAX);
row=i;
col=j;
}
}
//Determine whether the number is the smallest number in the same column
int flag=0;
for(int i=0;i<m;i++){
if(a[i][col]<a[row][col]){
flag++;//If the flag variable becomes 1, it means that the number is not the smallest number in the same column
MIN=a[i][col];
row=i;
}
flag=0;
}
if(flag==0){
printf("Array[%d][%d]=%d",row,col,a[row][col]);
break;
}
}
return 0;
}
Level 4: Binary Search
Title description: Number n integers sorted from small to large (n<1000000) from 1 to n, and an integer m to be searched, please use the dichotomy method to search.
Input The input consists of 3 lines, the first line contains integer n, the second line contains n integers separated by spaces, and the third line contains integer m.
Output If the integer m can be found in the sequence, output the number (if there are multiple numbers, return the smallest number), if not, output None.
#include<stdio.h>
int main(void)
{
/*********Begin*********/
int m,n,t;
scanf("%d",&n);
int a[n];
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}
//to sort
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(a[i]>a[j]){
t=a[i];
a[i] = a[j];
a[j] = t;
}
}
}
scanf("%d",&m);
int low=0;
int high=n-1;
while(1){
int mid=(low+high)/2;
if(a[mid]==m){
printf("%d",mid+1);
break;
}else if(a[mid]<m){
low=mid+1;
}else if(a[mid]>m){
high=mid-1;
}
if(low>high){
printf("None");
break;
}
}
return 0;
}
Level 3: Calculate the maximum value of the elements in the array and the subscript values of the rows and columns where they are located
Title description: According to the following function prototype programming, input a two-dimensional array with m rows and n columns from the keyboard, and then calculate the maximum value of the elements in the array and the subscript values of the rows and columns where they are located. Among them, the values of m and n are input by the user's keyboard. Neither m nor n is known to exceed 10.
enter
Enter the size of the array: "%d,%d" Enter the elements in the array below.
output
Output format: Array size input prompt information: "Input m, n:" Array element input prompt information: "Input %d*%d array: " Output format: "max=%d, row=%d, col=%d "
#include<stdio.h>
int main(void)
{
/*********Begin*********/
int m,n,col,row;
printf("Input m, n:");
scanf("%d,%d",&m,&n);
int a[m][n];
printf("Input %d*%d array:",m,n);
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
scanf("%d",&a[i][j]);
}
}
//output
int max=a[0][0];
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(a[i][j]>=max){
max= a[i][j];
row=i+1;
col=j+1;
}
}
}
printf("\nmax=%d, row=%d, col=%d",max,row,col);
return 0;
}
Level 2: Finding Integers
Title description: Given a sequence containing n integers, ask what is the first occurrence of the integer a in the sequence.
enter
The first line contains an integer n. The second line contains n non-negative integers, which is a given sequence, and each number in the sequence is not greater than 10000. The third line contains an integer a, which is the number to be searched.
output
If a appears in the sequence, output its first occurrence position (the position is numbered from 1), otherwise output -1.
Sample input:
6
1 9 4 8 3 9
9
Sample output:
2
Tip: Data size and conventions. 1 <= n <= 1000
#include<stdio.h>
int main(void)
{
/*********Begin*********/
int a[10],n,b,c=0;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}
scanf("%d",&b);
for(int i=0;i<n;i++){
if(a[i]==b){
c++;
printf("%d",i+1);
break;
}
}
if(c!=0){
}else{
printf("-1");
}
return 0;
}
Level 1: Sorting Problems
The task of this level: Arrange the ten numbers in descending order.
Input Enter ten integers.
Output Output the ten numbers in descending order.
#include<stdio.h>
int main(void)
{
/*********Begin*********/
int a[10],t;
for(int i=0;i<10;i++){
scanf("%d",&a[i]);
}
for(int i=0;i<10;i++){
for(int j=i+1;j<10;j++){
if(a[i]<a[j]){
t=a[i];
a[i]=a[j];
a[j]=t;
}
}
}
for(int m=0;m<10;m++){
printf("%d ",a[m]);
}
return 0;
}