意甲冠军：求0-B见面<=F[A]所有可能的

思维：数字DP，内存搜索

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;int A, B;int dp[20][200000];int bit[20];int dfs(int cur, int num, int flag) {	if (cur == -1)		return num >= 0;	if (num < 0)		return 0;	if (!flag && dp[cur][num] != -1)		return dp[cur][num];  	int ans = 0;	int end = flag?
        
       
      
     

bit[cur]:9; for (int i = 0; i <= end; i++) ans += dfs(cur-1, num-i*(1<<cur), flag&&i==end); if (!flag) dp[cur][num] = ans; return ans; } int F(int x) { int tmp = 0; int len = 0; while (x) { tmp += (x%10)*(1<<len); len++; x /= 10; } return tmp; } int cal() { int len = 0; while (B) { bit[len++] = B%10; B /= 10; } return dfs(len-1, F(A), 1); } int main() { int t; int cas = 1; scanf("%d", &t); memset(dp, -1, sizeof(dp)); while (t--) { scanf("%d%d", &A, &B); printf("Case #%d: %d\n", cas++, cal()); } return 0; }