ANARC08E spoj solution

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HII GUYS this is easy problem....... LOGIC:: when you want to arrange n different thing at n places so total number of ways=n! but when things are not different i.e let x+y=n it means x things are of one type and y things are of another type then total no ways to arrange this things at n places is=(x+y)!/(x! * y!) IN this problem total number of possibility also equal to (x+y)!/(x! * y!) m-1 you can direct do this MY AC PYTHON SOLUTION IS: if you have any problem plz comment me below import sys while(1): x,y=map(int,sys.stdin.readline().split()) if(x==-1 and y==-1): break fact1=1 fact2=1 fact3=1 for i in range(x): fact1=fact1*(i+1) for i in range(y): fact2=fact2*(i+1) for i in range(x+y): fact3=fact3*(i+1) ans=fact3/(fact1*fact2) if(ans==(x+y)): print "%d+%d=%d"%(x,y,(x+y)) else: print "%d+%d!=%d"%(x,y,(x+y)) m-2 you can this do by DP also there is one formula for binomial coefficient c(n,r)=c(n-1,r)+c(n-1,r-1) c(n,r)=c(n-r,r) here n=x+y so from 1st formula c(X+Y,X)=c(X+Y-1,X)+c(X+Y-1,X-1) now by using 2nd formula c(X+Y,X)=c(Y-1,X)+c(Y,X-1) my AC DP C++ solution is::if you have any problem plz comment me below...... #include<bits/stdc++.h> using namespace std; long long ans[11][11]; void pre() { long long i,j; for(i=0;i<11;i++) { for(j=0;j<11;j++) { if(i==0 || j==0) ans[i][j]=1; else ans[i][j]=ans[i-1][j]+ans[i][j-1]; } } } int main() { int a,b; pre(); while(1) { cin>>a>>b; if(a==-1 && b==-1) break; if((a+b)==ans[a][b]) printf("%d+%d=%d\n",a,b,a+b); else printf("%d+%d!=%d\n",a,b,a+b); } return 0; }