题目链接：uva
11645 - Bits

题目大意：给出n，问从0到n这n+1个数种，数的二进制情况下，有多少11存在。

解题思路：和uva 11038一个类型的题目，只是这道题目是对于二进制下的情况。而且高精度部分可以用两个long long数解决。

#include <cstdio>
#include <cstring>

typedef long long ll;
const int N = 100;
const ll M = 1e13;

ll bit (int k) {
    return (ll)1<<k;
}

void add (ll& p, ll& q, ll a, ll b) {
    if (a <= 0 || b <= 0)
        return;

    q += a * b;
    p += q/M;
    q %= M;
}

void solve (ll n) {

    int j = 0;
    ll left = n/2, right = 0;
    ll p = 0, q = 0;

    while (left) {

        left /= 2;
        add(p, q, left, bit(j));
        if ((n&bit(j)) && (n&bit(j+1)))
            add(p, q, 1, right+1);

        if (n&bit(j))
            right |= bit(j);
        j++;
    }

    if (p) {
        printf("%lld", p);
        printf("%013lld\n", q);
    } else {
        printf("%lld\n", q);
    }
}

int main () {
    ll n;
    int cas = 1;
    while (scanf("%lld", &n) == 1 && n >= 0) {
        printf("Case %d: ", cas++);
        solve(n);
    }
    return 0;
}

uva 11645 - Bits(计数问题+高精度)