poj 3252 Round Numbers 数位dp

题目链接

找一个范围内二进制中0的个数大于等于1的个数的数的数量。基础的数位dp

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define mem1(a) memset(a, -1, sizeof(a))
 4 int digit[50], dp[50][50][50], len;
 5 int dfs(int len, int num0, int num1, int f, int first) {        //first记录前面是否全部为0
 6     if(!len) {
 7         return num0>=num1;
 8     }
 9     if(!f&&!first&&dp[len][num0][num1]!=-1)
10         return dp[len][num0][num1];
11     int ret = 0, maxx = f?digit[len]:1;
12     for(int i = 0; i<=maxx; i++) {
13         if(i == 0) {
14             ret += dfs(len-1, first?0:num0+1, num1, f&&i==maxx, first);
15         } else {
16             ret += dfs(len-1, num0, num1+1, f&&i==maxx, 0);
17         }
18     }
19     if(!f&&!first)
20         return dp[len][num0][num1] = ret;
21     return ret;
22 }
23 int cal(int n) {
24     len = 0;
25     while(n) {
26         digit[++len] = n&1;
27         n>>=1;
28     }
29     return dfs(len, 0, 0, 1, 1);
30 }
31 int main()
32 {
33     int a, b;
34     mem1(dp);
35     while(~scanf("%d%d", &a, &b)) {
36         printf("%d\n", cal(b)-cal(a-1));
37     }
38 }

 

来源：https://www.cnblogs.com/yohaha/p/5035646.html