Answer:
Is longer and repeats key points.
Explanation:
When you start a live broadcast, viewers are received every second, and those who join the broadcast long after it starts may have no idea what you're talking about if you don't repeat the key points every so often.
The duration of the broadcast is also important, the longer it lasts, the more likely it is to receive more viewers and you can include more content to keep them entertained.
Answer:
#include<stdio.h>
#include<conio.h>
int m=0,n=4;
int cal(int temp[10][10],int t[10][10])
{
int i,j,m=0;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(temp[i][j]!=t[i][j])
m++;
}
return m;
}
int check(int a[10][10],int t[10][10])
{
int i,j,f=1;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
if(a[i][j]!=t[i][j])
f=0;
return f;
}
void main()
{
int p,i,j,n=4,a[10][10],t[10][10],temp[10][10],r[10][10];
int m=0,x=0,y=0,d=1000,dmin=0,l=0;
clrscr();
printf("\nEnter the matrix to be solved,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&a[i][j]);
printf("\nEnter the target matrix,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&t[i][j]);
printf("\nEntered Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",a[i][j]);
printf("\n");
}
printf("\nTarget Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",t[i][j]);
printf("\n");
}
while(!(check(a,t)))
{
l++;
d=1000;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(a[i][j]==0)
{
x=i;
y=j;
}
}
//To move upwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=0)
{
p=temp[x][y];
temp[x][y]=temp[x-1][y];
temp[x-1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move downwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x+1][y];
temp[x+1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move right side
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x][y+1];
temp[x][y+1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move left
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=0)
{
p=temp[x][y];
temp[x][y]=temp[x][y-1];
temp[x][y-1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
printf("\nCalculated Intermediate Matrix Value :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",r[i][j]);
printf("\n");
}
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
a[i][j]=r[i][j];
temp[i][j]=0;
}
printf("Minimum cost : %d\n",d);
}
getch();
}
Explanation:
Answer:
for (char i='a'; i<='e'; i++){
for (char j='a'; j<='e'; j++){
cout << i<< j<< "\n";
}
}
Explanation:
The loop runs all characters from the inner while the outer holds one character, by doing so, a will be matched with a,b,c,... Like wise b,c,d,... and on it goes.
What do you mean did you refresh the page
Answer:
tryparse method
Explanation:
You can convert a string to a number by calling the TryParse method found on various numeric types (int, long, double, etc.), or by using methods in the System.Convert class.
