Uvalive 6884 great + SWERC = PORTO DFS Simulation

Topic Connection:

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&itemid=8&page=show_problem& problem=4896

We want to has a great swerc at Porto this year and we approached this challenge in several ways.
We even framed it as a word addition problem, similar to the classic Send+more=money, where
Each letter stands for a digit (0, 1, 2, ..., 8, 9) that makes the arithmetic operation correct. Inch
Word Additions di?erent Letters cannot be assigned the same digit and the leftmost letter in a word
Cannot be zero (0). In particular, a single letter term cannot is zero.
To solve this word addition problem we had to nd positive digits for G, S and P, and digits for R,
E, A, T, W, C, O, so, all letter have A di?erent digit and the sum is correct. It turns out of that,
Unlike the classical Send+more=money which have a single solution, Great+swerc=porto
Has six solutions.
T=7, e=3, w=9, G=1, a=0, p=4, s=2, C=8, r=6, o=5
T=7, e=3, w=9, g=2, a=0, p=4, S=1, C=8, r=6, o=5
T=8, E=5, W=1, g=3, a=7, p=9, s=6, c=4, r=0, o=2
T=8, E=5, W=1, g=6, a=7, p=9, s=3, c=4, r=0, o=2
T=9, E=5, w=2, G=1, A=8, p=7, s=6, c=4, r=0, o=3
T=9, E=5, w=2, g=6, A=8, p=7, S=1, c=4, r=0, o=3
Have more than one solution does do great+swerc=porto a good problem to solve
by hand, but it was still a piece of cake for a programer. Moreover, it gives us another reason to organize
Swerc again next year and, who knows, in years to come!
Given A word addition problem, compute the number of solutions (possibly zero)

Test instructions

Give you n a string ask if you can add 1~n-1 to get nth string

You can replace a letter with a number in 0~9.

A number can replace only one letter

Different letters will not exceed 10,n no more than 10

Exercises

For each of the number of searches

Be patient and write well.

#include <bits/stdc++.h>using namespacestd;Const intN = -, M = 1e2+Ten, mod = 1e9+7, INF =2e9;typedefLong Longll;Chars[ One][ One];intcan =0, n,v[n],num[n];voidDfsintDepintLastintPintNow ) {    intf = strlen (s[n]+1); if(Dep>= (f+1)) {        if(last==0) can++; return ; } f= strlen (s[p]+1); if(p!=N) {if((F-DEP +1) >=1) {            if(V[S[P][F-DEP +1]] != -1) {                if(F-DEP +1==1&& V[S[P][F-DEP +1]]==0)return ; DFS (Dep,last,p+1, now+v[s[p][(F-DEP +1)]]); }Else {                 for(intI=0; i<=9; i++) {                    if(Num[i])Continue; if(F-DEP +1==1&&i==0)Continue; Num[i]=1; v[s[p][(f-Dep +1)]] =i; DFS (Dep,last,p+1, now+i); v[s[p][(f-Dep +1)]] = -1; Num[i]=0; }            }        }ElseDFS (dep,last,p+1, now); }Else {        if(v[s[p][(F-DEP +1)]]!=-1) {            if((f-dep+1==1) &&v[s[p][(F-DEP +1)]]==0)return ; if((now+last)%Ten! = v[s[p][(F-DEP +1)]])return ; DFS (DEP+1, (Now+last)/Ten,1,0); }        Else {            if(num[(now+last)%Ten])return ; if(dep==f&& (now+last)%Ten==0) {return ;} v[s[p][(f-Dep +1)]] = (now+last)%Ten; num[( now+last)%Ten] =1; DFS (DEP+1, (Now+last)/Ten,1,0); num[( now+last)%Ten] =0; v[s[p][(f-Dep +1)]] = -1; }    }}intMain () { while(~SCANF ("%d",&N)) { for(intI=1; i<=n;i++) scanf ("%s", s[i]+1); intMX =0;  for(intI=1; i<n;i++) {            intLen = strlen (s[i]+1); MX=Max (Mx,len); }        intMxx = strlen (s[n]+1); if(MX >mxx) {cout<<0<<Endl; Continue; } memset (num,0,sizeof(num)); memset (V,-1,sizeof(v)); Can=0; DFS (1,0,1,0); cout<<can<<Endl; }}

Uvalive 6884 great + SWERC = PORTO DFS Simulation