思路：用d[i][a][b][c][is]表示當前到了第i位， 三個數的i位分別是a,b,c， 是否有進位 ， 的方法數。
細節參見代碼：
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) using namespace std; typedef long long ll; typedef long double ld; const ld eps = 1e-9, PI = 3.1415926535897932384626433832795; const int mod = 1000000000 + 7; const int INF = 0x3f3f3f3f; // & 0x7FFFFFFF const int seed = 131; const ll INF64 = ll(1e18); const int maxn = 15; int T,n,m,len,vis[maxn][maxn][maxn][maxn][2],len1,len2,len3,kase = 0; char a[maxn],b[maxn],c[maxn],s[100]; ll d[maxn][maxn][maxn][maxn][2]; ll dp(int pos, int bb, int cc, int dd, int is) { ll& ans = d[pos][bb][cc][dd][is]; if(pos > len) return is == 0; if(vis[pos][bb][cc][dd][is] == kase) return ans; vis[pos][bb][cc][dd][is] = kase; ans = 0; if(a[pos] == '?' && b[pos] == '?') { for(int i = 0; i < 10; i++) { for(int j = 0; j < 10; j++) { if(pos == len1 && i == 0 && len1 != 1) continue; if(pos == len2 && j == 0&& len2 != 1) continue; int cur = i + j + is; int res = 0; if(cur >= 10) { cur -= 10; ++res; } if(c[pos] == '?' && !(pos == len3 && cur == 0 && len3 != 1)) ans += dp(pos+1,i, j,cur, res); else if(c[pos]-'0' == cur) ans += dp(pos+1, i,j,cur, res); } } } else if(a[pos] == '?') { for(int i = 0; i < 10; i++) { if(pos == len1 && i == 0&& len1 != 1) continue; int cur = i + b[pos]-'0' + is; int res = 0; if(cur >= 10) { cur -= 10; ++res; } if(c[pos] == '?' && !(pos == len3 && cur == 0 && len3 != 1)) ans += dp(pos+1,i,b[pos]-'0',cur, res); else if(c[pos]-'0' == cur) ans += dp(pos+1,i,b[pos]-'0',cur, res); } } else if(b[pos] == '?') { for(int i = 0; i < 10; i++) { if(pos == len2 && i == 0&& len2 != 1) continue; int cur = i + a[pos]-'0' + is; int res = 0; if(cur >= 10) { cur -= 10; ++res; } if(c[pos] == '?' && !(pos == len3 && cur == 0 && len3 != 1)) ans += dp(pos+1,a[pos]-'0',i,cur, res); else if(c[pos]-'0' == cur) ans += dp(pos+1,a[pos]-'0',i,cur, res); } } else { int cur = a[pos]-'0' + b[pos]-'0' + is; int res = 0; if(cur >= 10) { cur -= 10; ++res; } if(c[pos] == '?' && !(pos == len3 && cur == 0 && len3 != 1)) ans += dp(pos+1, a[pos]-'0',b[pos]-'0',cur, res); else if(c[pos]-'0' == cur) ans += dp(pos+1, a[pos]-'0',b[pos]-'0',cur, res); } return ans; } int main() { while(~scanf("%s",s+1)) { len = strlen(s+1); len1 = 0; len2 = 0; len3 = 0; int id = 0; for(int i = len; i >= 1; i--) { if(s[i] == '=' || s[i] == '+') { id++; continue; } if(id == 0) { c[++len3] = s[i]; } else if(id == 1) { b[++len2] = s[i]; } else { a[++len1] = s[i]; } } len = max(len1, max(len2, len3));//補全不足的， 減少代碼量 for(int i = len1+1; i <= len; i++) a[i] = '0'; for(int i = len2+1; i <= len; i++) b[i] = '0'; for(int i = len3+1; i <= len; i++) c[i] = '0'; ++kase; ll ans = dp(1, 0 , 0, 0, 0); printf("Case %d: %I64d\n",kase,ans); } return 0; }
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