Hello, Hello, everyone. I'm a self-motivated kid Vegetable pig
Gou huaisifangzhi , Where you can swim , Let's make progress together ！
label ： Fill in the blank , 2015, Provincial competition
Title Description
This topic is code completion and filling in the blanks , Please complete the source code given in the title , And copy it to the code box on the right , Select the corresponding compilation language （C/Java） Submit after . If the source language given in the title is not unique , Then you only need to select one of them to complete the submission . After copying, delete the underline in the blank part of the source code , Fill in your answer . If it fails to pass after submission , In addition to considering the mistakes in filling in the blanks , You should also pay attention to whether there are errors due to changes in the non blank parts after copying .
StringInGrid The function prints the specified string in a specified size grid .
The string is required to be horizontal 、 It's centered in both directions .
If the string is too long , Just cut off .
If you can't just Center , It can be a little bit left or a little bit up .
The following program implements this logic , Please fill in the missing code in the underlined section .
Source code
#include <stdio.h>
#include <string.h>
void StringInGrid(int width, int height, const char* s)
{
int i,k;
char buf[1000];
strcpy(buf, s);
if(strlen(s)>width-2) buf[width-2]=0;
printf("+");
for(i=0;i<width-2;i++) printf("-");
printf("+\n");
for(k=1; k<(height-1)/2;k++){
printf("|");
for(i=0;i<width-2;i++) printf(".");
printf("|\n");
}
printf("|");
printf("%*s%s%*s",__________________);
printf("|\n");
for(k=(height-1)/2+1; k<height-1; k++){
printf("|");
for(i=0;i<width-2;i++) printf(".");
printf("|\n");
}
printf("+");
for(i=0;i<width-2;i++) printf("-");
printf("+\n");
}
int main()
{
StringInGrid(10,4,"abcd123");
return 0;
}c++ solution ：
#include <stdio.h>
#include <string.h>
void StringInGrid(int width, int height, const char* s)
{
int i,k;
char buf[1000];
strcpy(buf, s);
if(strlen(s)>width-2) buf[width-2]=0;
printf("+");
for(i=0;i<width-2;i++) printf("-");
printf("+\n");
for(k=1; k<(height-1)/2;k++){
printf("|");
for(i=0;i<width-2;i++) printf(".");
printf("|\n");
}
printf("|");
printf("%*s%s%*s",0,"",s,1,"");
printf("|\n");
for(k=(height-1)/2+1; k<height-1; k++){
printf("|");
for(i=0;i<width-2;i++) printf(".");
printf("|\n");
}
printf("+");
for(i=0;i<width-2;i++) printf("-");
printf("+\n");
}
int main()
{
StringInGrid(10,4,"abcd123");
return 0;
}Summary of knowledge points ：
1. stay printf In the sentence * Represents a more flexible control domain If written printf("%6d", 123); %6s Also domain width ;
2.printf("%*s",3,"") Represents the output of three spaces .
3. Use %s, Indicates that the specific field width here is determined by the following arguments , Such as printf("%s",6, “abc”) Is to put "abc" Put in a field with a width of 6 Right aligned in the space of .
label ： Fill in the blank , 2015, Provincial competition
Title Description
This topic is code completion and filling in the blanks , Please complete the source code given in the title , And copy it to the code box on the right , Select the corresponding compilation language （C/Java） Submit after . If the source language given in the title is not unique , Then you only need to select one of them to complete the submission . After copying, delete the underline in the blank part of the source code , Fill in your answer . If it fails to pass after submission , In addition to considering the mistakes in filling in the blanks , You should also pay attention to whether there are errors due to changes in the non blank parts after copying .
1,2,3...9 These nine numbers make up a fraction , The value is exactly 1/3, How to organize ？
The following program implements this function , Please fill in the missing code in the underlined part .
Source code ：
#include <stdio.h>
void test(int x[])
{
int a = x[0]*1000 + x[1]*100 + x[2]*10 + x[3];
int b = x[4]*10000 + x[5]*1000 + x[6]*100 + x[7]*10 + x[8];
if(a*3==b) printf("%d/%d\n", a, b);
}
void f(int x[], int k)
{
int i,t;
if(k>=9){
test(x);
return;
}
for(i=k; i<9; i++){
{t=x[k]; x[k]=x[i]; x[i]=t;}
f(x,k+1);
_________________
}
}
int main()
{
int x[] = {1,2,3,4,5,6,7,8,9};
f(x,0);
return 0;
}analysis ：
for(i = k; i < 9; i++){t = xk;xk = xi;xi = t; to flash back ;}
c++ solution ：
#include <stdio.h>
void test(int x[])
{
int a = x[0]*1000 + x[1]*100 + x[2]*10 + x[3];
int b = x[4]*10000 + x[5]*1000 + x[6]*100 + x[7]*10 + x[8];
if(a*3==b) printf("%d/%d\n", a, b);
}
void f(int x[], int k)
{
int i,t;
if(k>=9){
test(x);
return;
}
for(i=k; i<9; i++)
{
{t=x[k]; x[k]=x[i]; x[i]=t;}
f(x,k+1);
{t=x[k]; x[k]=x[i]; x[i]=t;}
}
}
int main()
{
int x[] = {1,2,3,4,5,6,7,8,9};
f(x,0);
return 0;
}label ： Fill in the blanks , 2015, Provincial competition
We all know ：1+2+3+ ... + 49 = 12251+2+3+...+49=1225
Now you are required to change two of the nonadjacent plus signs into multipliers , Make the result 20152015
such as ：
1+2+3+...+1011+12+...+2728+29+...+49 = 2015
1+2+3+...+10∗11+12+...+27∗28+29+...+49=2015
It's the answer that meets the requirements .
Please look for another possible answer , And submit the number to the left of the multiplier in front of you （ For example , Is to submit 1010）.
Be careful ： What you need to submit is an integer , Don't fill in any superfluous information .
answer ：
16
c++ solution ：
#include<bits/stdc++.h>
using namespace std;
int main()
{
for(int i=1;i<47;i++)
{
for(int j=i+2;j<49;j++)
{
if((i*(i+1)-(i+i+1)+(j*(j+1)-(j+j+1))==2015-1225))
{
cout<<i<<" "<<j<<endl;
}
}
}
}python practice ：
for i in range(1,47): for j in range(i+2,49): if(i*(i+1)-(i+i+1)+j*(j+1)-(j+j+1)==2015-1225): print(i,j)
Ideas ：
1+2+3+ ... + 49 = 12251+2+3+...+49=1225
1+2+3+...+10∗11+12+...+27∗28+29+...+49=2015
The sum of the remaining items remains unchanged , Enumerating the sum of multiplication products equals to the difference between the answers .
If there is a better solution and its ideas , Welcome to discuss .
