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
#include<bits/stdc++.h>
using namespace std;
#define mem1(a) memset(a, -1, sizeof(a))
int digit[], dp[][][], len;
int dfs(int len, int num0, int num1, int f, int first) { //first记录前面是否全部为0
if(!len) {
return num0>=num1;
}
if(!f&&!first&&dp[len][num0][num1]!=-)
return dp[len][num0][num1];
int ret = , maxx = f?digit[len]:;
for(int i = ; i<=maxx; i++) {
if(i == ) {
ret += dfs(len-, first?:num0+, num1, f&&i==maxx, first);
} else {
ret += dfs(len-, num0, num1+, f&&i==maxx, );
}
}
if(!f&&!first)
return dp[len][num0][num1] = ret;
return ret;
}
int cal(int n) {
len = ;
while(n) {
digit[++len] = n&;
n>>=;
}
return dfs(len, , , , );
}
int main()
{
int a, b;
mem1(dp);
while(~scanf("%d%d", &a, &b)) {
printf("%d\n", cal(b)-cal(a-));
}
}