UVALive 6918And Or（位运算）

题意:

求1≤L≤R≤1018区间的所有的数的AND和OR值

分析:

OR看1,AND看0
lowbit makes lowest 1 to 0 for AND
lowbit makes lowest 0 to 1(except trailing zeros) for OR
所以对于OR的话后缀零要特判一下

代码:

////  Created by TaoSama on 2015-12-08//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        LL l, r; scanf("%lld%lld", &l, &r);//      LL a = l, b = l;//      for(LL i = l; i <= r; ++i) a |= i, b &= i;        LL ansAnd = l, ansOr = l;        //trailing zeros for OR        for(int i = 0; i < 64; ++i) {            if(l >> i & 1) break;            LL tmp = l + (1LL << i);            if(tmp <= r) ansOr |= tmp;        }        //lowbit makes lowest 1 to 0 for AND        //lowbit makes lowest 0 to 1(except trailing zeros) for OR        while(l <= r) {            ansOr |= l;            ansAnd &= l;            l += l & -l;        }        printf("Case %d: %lld %lld\n", ++kase, ansOr, ansAnd);//      printf("Case %d: %lld %lld\n\n", kase, a, b);    }    return 0;}