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DP[len][k][i][j] 再第len位,第一个数len位为i,第二个数len位为j,和的第len位为k
每一位能够从后面一位转移过来,能够进位也能够不进位
3+??
1=44?” can be completed to “123+321=444” , “143+301=444” and many other possible solutions. Your job is to determine the number of different possible solutions.
7+1?=1? ?1+?1=22
Case 1: 3 Case 2: 1HintThere are three solutions for the first case: 7+10=17, 7+11=18, 7+12=19 There is only one solution for the second case: 11+11=22 Note that 01+21=22 is not a valid solution because extra leading zeros are not allowed.
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <stack> using namespace std; typedef long long int LL; char cpp[200]; int a[200],len1,b[200],len2,c[200],len3; LL dp[20][20][20][20]; int main() { int cas=1; while(cin>>cpp) { len1=len2=len3=0; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); int n=strlen(cpp); int i; stack<char> stk; for(i=0;i<n;i++) { if(cpp[i]=='+') { while(!stk.empty()) { char c=stk.top(); stk.pop(); if(c!='?') a[len1++]=c-'0'; else a[len1++]=-1; } i++; break; } stk.push(cpp[i]); } for(;i<n;i++) { if(cpp[i]=='=') { while(!stk.empty()) { char c=stk.top(); stk.pop(); if(c!='?
') b[len2++]=c-'0'; else b[len2++]=-1; } i++; break; } stk.push(cpp[i]); } for(;i<n;i++) stk.push(cpp[i]); while(!stk.empty()) { char cc=stk.top(); stk.pop(); if(cc!='?') c[len3++]=cc-'0'; else c[len3++]=-1; } for(int i=len1-1;i>0;i--) if(a[i]==0) len1--; else break; for(int i=len2-1;i>0;i--) if(b[i]==0) len2--; else break; for(int i=len3-1;i>0;i--) if(c[i]==0) len3--; else break; memset(dp,0,sizeof(dp)); ///len==0 for(int i=0;i<=9;i++) { if(a[0]==-1||a[0]==i) for(int j=0;j<=9;j++) { if(b[0]==-1||b[0]==j) for(int k=0;k<=9;k++) if(c[0]==-1||c[0]==k) { if(k==(i+j)%10) dp[0][k][i][j]=1; } } } ///len=1... for(int len=1;len<len3;len++) { for(int i=0;i<=9;i++) { if(len==len1-1&&i==0) continue; if(len>=len1&&i!=0) continue; if(a[len]==-1||a[len]==i) for(int j=0;j<=9;j++) { if(len==len2-1&&j==0) continue; if(len>=len2&&j!=0) continue; if(b[len]==-1||b[len]==j) for(int k=0;k<=9;k++) { if(len==len3-1&&k==0) continue; if(((i+j)%10!=k)&&((i+j+1)%10!=k)) continue; if(c[len]==-1||c[len]==k) { ///没有进位 if((i+j)%10==k) { for(int ii=0;ii<=9;ii++) for(int jj=0;jj<=9;jj++) for(int kk=0;kk<=9;kk++) { if((ii+jj==kk)||(ii+jj+1==kk)) dp[len][k][i][j]+=dp[len-1][kk][ii][jj]; } } ///有进位 if((i+j+1)%10==k) { for(int ii=0;ii<=9;ii++) for(int jj=0;jj<=9;jj++) for(int kk=0;kk<=9;kk++) { if(((ii+jj>=10)&&(ii+jj)%10==kk)||((ii+jj+1>=10)&&(ii+jj+1)%10==kk)) dp[len][k][i][j]+=dp[len-1][kk][ii][jj]; } } } } } } } LL ans=0; int mx=max(len1,max(len2,len3)); for(int i=0;i<=9;i++) for(int j=0;j<=9;j++) for(int k=0;k<=9;k++) if((i+j==k)||(i+j+1==k)) { if(mx==1&&i+j!=k) continue; ans+=dp[mx-1][k][i][j]; } cout<<"Case "<<cas++<<": "<<ans<<endl; memset(cpp,0,sizeof(cpp)); } return 0; }
HDOJ 4249 A Famous Equation DP
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原文地址:http://www.cnblogs.com/mengfanrong/p/5161644.html