题目链接:点击打开链接
思路:数电中, 写真值表的时候, 我们是按照一定的规则写的。 按照这个规则, 我们可以知道比一个数小的数中, 每一位有多少个1。 所以, 对于每一位, 进位数就是1的个数除以2加上上一次进位的个数再除以2向下取整。
细节参见代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#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 double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int seed = 131;
const ll INF64 = ll(1e18);
int n, k;
int a, b;
ll solve(int u, int v) {
int cur = 2, maxv ;
int lastjinwei = 0;
ll ans = 0;
for(int i = 30; i >= 0; i--) {
if(v & (1<<i)) {
maxv = i;
break;
}
}
for(int i = 0; i <= maxv; i++) {
int cnt = 0;
if((v+1)%cur == 0) {
int dui = (v+1)/cur;
int _num = (1<<i);
cnt = dui * _num;
}
else {
int dui = (v+1)/cur;
int _num = (1<<i);
cnt = dui * _num;
int res = (v+1) - dui * cur;
if(res <= cur/2) ;
else {
cnt += res - cur/2;
}
}
int cnt2 = 0;
if((u+1)%cur == 0) {
int dui = (u+1)/cur;
int _num = (1<<i);
cnt2 = dui*_num;
}
else {
int dui = (u+1)/cur;
int _num = (1<<i);
cnt2 = dui * _num;
int res = (u+1) - dui * cur;
if(res <= cur/2) ;
else {
cnt2 += res - cur/2;
}
}
cnt -= cnt2;
cnt += lastjinwei;
ans += cnt/2;
lastjinwei = cnt/2;
cur *= 2;
}
while(true) {
ans += lastjinwei/2;
lastjinwei /= 2;
if(lastjinwei <= 1) break;
}
return ans;
}
int main() {
// ios::sync_with_stdio(false);
while(~scanf("%d%d", &a, &b)) {
printf("%I64d\n", solve(a-1, b));
}
return 0;
}