Water + Earth = Melon

Computer Science Level 4

WATER +EARTH ‾ MELON \Large{\begin{matrix} \text{WATER} \\ \underline { \text{+EARTH} } \\ \text{MELON} \end{matrix}}

If above given is a cryptogram such that each distinct letter represents distinct single digit numbers (i.e. 0,1, 2, 3, ...., 9). Find the sum of all values of MELON \text{MELON} . Note:- W , E , M ≠ 0 W,E,M \neq 0


The answer is 652560.

Department 8 by Department 8 , 27, Israel

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3 solutions

It takes a long time 

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#include<bits/stdc++.h>
using namespace std;

int main(){
int w,a,t,e,r,m,l,o,n,h;
long long shell,elf,sale,rest;
long long bill=0;

w=1; a=0; t=0; e=1; r=0;
while(w<10)  {
    while(a<10){
        while(t<10){
            while(e<10){
                while(r<10){
                    for(m=1;m<10;m++){
                        for(l=0;l<10;l++){
                            for(o=0;o<10;o++){
                                for(n=0;n<10;n++){
                                        for(h=0;h<10;h++){
if((w!=a) && (w!=t) && (w!=e) && (w!=r) && (w!=m) && (w!=l) && (w!=o) && (w!=n) && (w!=h) &&
   (a!=t) && (a!=e) && (a!=r) && (a!=m) && (a!=l) && (a!=o) && (a!=n) && (a!=h) &&
   (t!=e) && (t!=r) && (t!=m) && (t!=l) && (t!=o) && (t!=n) && (t!=h) &&
   (e!=r) && (e!=m) && (e!=l) && (e!=o) && (e!=n) && (e!=h) &&
   (r!=m) && (r!=l) && (r!=o) && (r!=n) && (r!=h) && 
   (m!=l) && (m!=o) && (m!=n) && (m!=h) &&
   (l!=o) && (l!=n) && (l!=h) &&
   (o!=n) && (o!=h) &&
   (n!=h)){
                        shell=w*10000+a*1000+t*100+e*10+r; elf=e*10000+a*1000+r*100+t*10+h; sale=m*10000+e*1000+l*100+o*10+n;
                        if((shell+elf)==sale) bill+=sale;
                                }
                                }
                                }
                    }
                    }
                }
                r++;}
            e++;r=0;}
        t++;e=1;}
    a++;t=0;}
w++;a=0;}   

cout<<bill<<endl; //OUTPUT=652560

}

about 171 sec.

1 Helpful 0 Interesting 0 Brilliant 1 Confused

Good but not better than mine :P

Zeeshan Ali - 5 years, 3 months ago

Log in to reply

"while" faster than "for", I think :/

Resha Dwika Hefni Al-Fahsi - 5 years, 3 months ago

If you want to find all the values, you don't need to wait for that long; here's my python code: 

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for w,a,t,e,r,h,m,l,o,n in permutations('0123456789',10):
  if '0' in (w,e,m):
    continue
  if int(w+a+t+e+r)+int(e+a+r+t+h)==int(m+e+l+o+n):
    print(m+e+l+o+n)

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Zeeshan Ali
Jan 16, 2016

Well... It was so boring to wait so long for the result but I did it... 

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#include<iostream>
using namespace std;
int main(){
    long long int sum=0;
    for (int w=1; w<10; w++){
    for (int a=0; a<10; a++){
    for (int t=0; t<10; t++){
    for (int e=1; e<10; e++){
    for (int r=0; r<10; r++){
    for (int h=0; h<10; h++){
    for (int m=1; m<10; m++){
    for (int l=0; l<10; l++){
    for (int o=0; o<10; o++){
    for (int n=0; n<10; n++){
        if (!((w+e+(a+a+(t+r+(e+t+(r+h)/10)/10)/10)/10)/10)){
            if (w!=a && w!=t && w!=e && w!=r && w!=h && w!=m && w!=l && w!=o && w!=n){
                if (a!=t && a!=e && a!=r && a!=h && a!=m && a!=l && a!=o && a!=n){
                    if (t!=e && t!=r && t!=h && t!=m && t!=l && t!=o && t!=n){
                        if (e!=r && e!=h && e!=m && e!=l && e!=o && e!=n){
                            if (r!=h && r!=m && r!=l && r!=o && r!=n){
                                if (h!=m && h!=l && h!=o && h!=n){
                                    if (m!=l && m!=o && m!=n){
                                        if (l!=o && l!=n){
                                            if (o!=n){                              
                                                if (w!=a && w!=t && w!=e && w!=r && w!=h && w!=m && w!=l && w!=o && w!=n){
                                                    int A[5];
                                                    A[0]=(r+h)%10;
                                                    A[1]=(e+t+(r+h)/10)%10;
                                                    A[2]=(t+r+(e+t+(r+h)/10)/10)%10;
                                                    A[3]=(a+a+(t+r+(e+t+(r+h)/10)/10)/10)%10;
                                                    A[4]=(w+e+(a+a+(t+r+(e+t+(r+h)/10)/10)/10)/10)%10;
                                                    if (A[0]==n && A[1]==o && A[2]==l && A[3]==e && A[4]==m){
                                                        long long int term=m*10000+e*1000+l*100+o*10+n;
                                                        sum+=term;
                                                    }
                                                }
                                            }
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    }
    }
    }
    }
    }
    }
    }
    }
    }
    cout<<sum<<endl;
    return 0;
}

0 Helpful 0 Interesting 0 Brilliant 0 Confused 

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for c in range(1,10):
    for h in range(10):
        if h!=c:
            for a in range(10):
                if a not in [h,c]:
                    for r in range(1,10):
                        if r not in [c,h,a]:
                            for y in range(10):
                                if y not in [c,h,a,r]:  
                                    for s in range(1,10):
                                        if s not in [c,h,a,r,y]:    
                                            for e in range(10):
                                                if e not in [c,h,a,r,y,s]:
                                                    for m in range(10):
                                                        if m not in [c,h,a,r,y,s,e]:
                                                            for u in range(10):
                                                                if u not in [c,h,a,r,y,s,e,m]:
                                                                    for b in range(10):
                                                                        if b not in [c,h,a,r,y,s,e,m,u]:
                                                                            if (10000*(c+r-s)+1000*(h+h-r)+100*(a+y-e)+10*(r+a-m)+y+b-u==0):
                                                                                print s,r,e,m,u

did not take much time.

Soumava Pal - 5 years, 3 months ago

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