@sam_p30 I had the same logic. I made all possible combinations from the numbers, and then printed maximum, but I got WA.
Here is my code in c++: CodeChef: Practical coding for everyone
1 Like
You are adding all digits like unit digits
1 Like
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main() {
int t;
cin>>t;
while(t--) {
int a,b;
cin>>a>>b;
int mx=a+b;
int a1=a%10;
int a2=a/10;
int b1=b%10;
int b2=b/10;
if(b2) mx=max(mx,a2*10+b2+a1*10+b1);
if(a2) mx=max(mx,b1*10+a1+b2*10+a2);
cout<<mx<<endl;
}
return 0;
}
//1-mx=max(mx,a210+b1+b210+a1);
//2-if(a2 && b2) mx=max(mx,b210+a1+a210+b1);
you can also remove these conditions from your code
becoz-
1- statement is swapping only ones’s digit of both the numbers,which would not affect the sum
eg - 45+76=121 and 75+46=121
2-statement is swapping only ten’s digit of both numbers ,if they both are double digit numbers,which would again wont affect the sum as even if both are single digit or one of them is single and other is double digit it WONT AFFECT the sum
eg-34+52=86 and 32+54=86 both are equal
4 Likes
please can anyone point me out why it gives WA @taran_1407
#include<bits/stdc++.h>
using namespace std;
int main()
{
int tc,x;
cin>>tc;
for(x=0;x<tc;x++){
int a,b,i,max1=0;
int sum[5],dig[4];
cin>>a>>b;
dig[0]=a/10;
dig[1]=a%10;
dig[2]=b/10;
dig[3]=b%10;
sum[0]=a+b;
max1=sum[0];
sum[1]=(((dig[0]*10)+dig[3])+((dig[2]*10)+dig[1]));
sum[2]=(((dig[0]*10)+dig[2])+((dig[1]*10)+dig[3]));
sum[3]=(((dig[2]*10)+dig[1])+((dig[0]*10)+dig[3]));
sum[4]=(((dig[3]*10)+dig[1])+((dig[2]*10)+dig[0]));
for(i=1;i<5;i++){
if(max1<sum[i]){
max1=sum[i];
}
}
cout<<max1<<"\n";
}
return 0;
}1
99 9
Actual output :- 189
Expected Output :- 108
You are not supposed to find sum[4] for (9,99) and sum[2] for reverse of this test case(99,9) for this kind of test cases. Because test case can be interpreted like(99,09) and if you swap dig[3] with dig[2] i.e (90,99). since zero does not hold any value before a number(for 2 digit number) and add them surely it is going to give wa.
For single digit test cases like (1,3) and (3,1) you should not calculate sum[4] and sum[2].
4 Likes
Hi everyone . I want help from the pro coders here, can you check whats wrong with my python code for this question, it will be a big help .
please check why my code is giving WA …
#include<iostream.h>
#include<bits/stdc++.h>
using namespace std;
int main() {
// your code goes here
int t;
cin>>t;
while(t–>0)
{
int a,b;
cin>>a>>b;
int sum=0;
sum=a+b;
int a1=0,b1=0,b2=0,a2=0;
a1=a/10;
a2=a%10;
b1=b/10;
b2=b%10;
if(a1==0 && b1==0);
if(a1==0 && b1!=0)
{
if((a2 * 10 + b1+ b2 )>sum)
sum=a2* 10 + b1+ b2 ;
}
else if(b1==0 && a1!=0)
{
if(( b2 * 10 + a1 + a2 )>sum)
{
sum=b2*10 + a1 + a2;
}
}
else
{
if((a1*10+ a2*10 + b1 +b2)>sum)
{
sum=(a1+a2)*10 + b1 +b2;
}
else if((b1*10+b2*10 + a1+ a2)>sum)
{
sum=(b1+b2)*10 + a1+ a2;
}
}
cout<< sum <<endl;
}
return 0;
}
can someone help me ? why it doesnt accept my solution?
#include<stdio.h>
int main(){
int A,B,i=2,j=2,array[3]={},array2[3]={},a,T,T2,T3,k;
printf(“How many test do you want?\n”);
scanf("%d",&T2);
if(T2>=1 && T2<=1000){
do{
for(i=0;i<3;i++){
array[i]=0;
array2[i]=0;
}
printf("enter 2 number.\n");
scanf("%d %d",&A,&B);
if(A<100&& A>0 && B>0 && B<100){
T=A;i=2;j=2;
while(T!=0){
a=T%10;
array[i]=a;
i--;
T=T/10;
}
T=B;
while(T!=0){
a=T%10;
array2[j]=a;
j--;
T=T/10;
}
if(array[2]>array2[1] && array[2]>array2[2] && array2[1]!=0 ){
T3=array[2];
array[2]=array2[1];
array2[1]=T3;
}
else if(array[1]<array2[2] && array2[2]>array[2] && array[1]!=0){
T3=array[1];
array[1]=array2[2];
array2[2]=T3;
}
else if(array[1]<array2[2] && array[1]!=0){
T3=array[1];
array[1]=array2[2];
array2[2]=T3;
}
else if(array[2]>array2[1] && array2[1]!=0){
T3=array[2];
array[2]=array2[1];
array2[1]=T3;
}
array[0]=array[1]*10+array[2];
array2[0]=array2[1]*10+array2[2];
printf("%d\n",array[0]+array2[0]);
k++;
}
}while(k!=T2);
}
return 0;
}
Can someone please tell in which test cases my code fails? It is showing WA
cook your dish here
for _ in range(int(input())):
a,b=input().split()
unit1=int(a[-1])
unit2=int(b[-1])
tens1=(int(a[0]) if len(a)==2 else 0)
tens2=(int(b[0]) if len(b)==2 else 0)
if (unit1>tens2 and tens2!=0):
unit1,tens2=tens2,unit1
elif (unit2>tens1 and tens1!=0):
unit2,tens1=tens1,unit2
print(((tens1+tens2)*10)+(unit2+unit1))why my solution is not accepted?
i am checking manual test cases and getting correct answers but judge is not accepting the solution.please check my approach …
#include
using namespace std;
int main() {
// your code goes here
int t;
cin>>t;
for(int i=0;i<t;i++)
{
int a,b;
cin>>a>>b;
if(a<10&&b<10)
{
cout<<a+b<<endl;
}
else if(a<10&&b>=10)
{
int c=(b/10);
int d=(a*10)+(b%10);
int e=b%10;
int f=b-(b%10)+a;
if(a+b>c+d && a+b>e+f)
cout<<a+b<<endl;
else if(c+d>a+b && c+d>e+f)
cout<<c+d<<endl;
else
cout<<e+f<<endl;
}
else if(b<10&&a>=10)
{
int c=(a/10);
int d=(b*10)+(a%10);
int e=a%10;
int f=a-(a%10)+b;
if(b+a>c+d && b+a>e+f)
cout<<b+a<<endl;
else if(c+d>b+a && c+d>e+f)
cout<<c+d<<endl;
else
cout<<e+f<<endl;
}
else
{
int c=(a%10)+(b-b%10);
int d=(b%10)+(a-a%10);
int e=(a%10)+(b%10)*10;
int f=(b-b%10)+a/10;
int g=(a-a%10)+b/10;
int h=(b%10)+(a%10)*10;
int i=(a-a%10)+b%10;
int j=(b-b%10)+a%10;
if(a+b>c+d&&a+b>e+f&&a+b>g+h&&a+b>i+j)
cout<<a+b<<endl;
else if(c+d>a+b&&c+d>e+f&&c+d>g+h&&c+d>i+j)
cout<<c+d<<endl;
else if(e+f>c+d&&e+f>a+b&&e+f>g+h&&e+f>i+j)
cout<<e+f<<endl;
else if(g+h>c+d&&g+h>a+b&&g+h>e+f&&g+h>i+j)
cout<<g+h<<endl;
else
cout<<i+j<<endl;
}
}
return 0;
}
Can anyone explain what is wrong in my approach?
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int llu;
#define gcd(a,b) __gcd(a,b)
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while(t--){
string a,b;
cin>>a>>b;
int mina,minb;
int indexa,indexb;
if(a.size()==1){
mina=a[0];
indexa=0;
}
else{
if(a[0]<=a[1]){
mina=a[0];
indexa=0;
}
else{
mina=a[1];
indexa=1;
}
}
if(b.size()==1){
minb=b[0];
indexb=0;
}
else{
if(b[0]<=b[1]){
minb=b[0];
indexb=0;
}
else{
minb=b[1];
indexb=1;
}
}
//cout<<mina-48<<" "<<minb-48<<" "<<indexa<<" "<<indexb<<endl;
a[indexa]=minb;
b[indexb]=mina;
int x,y;
if(a.size()==1){
x=a[0]-48;
}
else{
x=(a[0]-48)*10+(a[1]-48);
}
if(b.size()==1){
y=b[0]-48;
}
else{
y=(b[0]-48)*10+(b[1]-48);
}
cout<<x+y<<endl;
}
return 0;
}Can somebody help, why is my code giving wrong answer.
https://www.codechef.com/viewsolution/32158818
Really, it’s good to know.
why do we have to put the if(b2) and if(a2) condition without that also well get the same answer na?
You have to consider all the four cases. That are:
if a and b are numbers:
1st case : a>=10 and b>=10
2nd case : a>=10 and b<10
3rd case : a<10 and b>=10
4th case : a<10 and b<10
Here is the link to my code : CodeChef: Practical coding for everyone
Hope it helps…
@taran_1407 can you please explain the logic for the bonus part of this question… i.e.,
Can you solve it without explicitly doing swaps and then comparing?
I m not able to find solution for this. Please guide…
thank you so much for your testcase.
can anyone explain the setter solution
mx=max(mx,a210+b1+b210+a1);
if(b2) mx=max(mx,a210+b2+a110+b1);
if(a2) mx=max(mx,b110+a1+b210+a2);
if(a2 && b2) mx=max(mx,b210+a1+a210+b1);
i am not getting the logic behind this code