题意：给定一个数 n，求 0 ~ n，中二进制表示中连续两个 1 出现的次数。
析：枚举连续的两个 1，从低位向高位进行枚举，然后前可以是任意数，后面也是任意的，如果 n 正好是 11 还要另算，举个例子。
10110，假设现在枚举第 2 位和第 3 位，那么出现的次次数就是前面的 10，还有第一位是任意的，所以就有 10 = 2 * 2 = 4 种，而且正好第 2 位和第 3 位是 1，那么对于第一位也是是随便的，再加上 2。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1500 + 50;
const int maxm = 1e6 + 10;
const LL mod = 1000000000000000LL;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x; scanf("%d", &x); return x; }
LL p, q;
void add(LL x){
q += x;
p += q / mod;
q %= mod;
}
int main(){
int kase = 0;
LL n;
while(cin >> n && n >= 0){
p = q = 0;
LL m = 1LL, t = n;
while(n){
add((n>>2) * m);
if((n&3) == 3) add((t&m-1) + 1);
n >>= 1;
m <<= 1;
}
printf("Case %d: ", ++kase);
if(p) printf("%lld%015lld\n", p, q);
else printf("%lld\n", q);
}
return 0;
}
原文：https://www.cnblogs.com/dwtfukgv/p/8969824.html