HDU 5491 The Next （二进制） 2015合肥网络赛

题意: 求数的二进制位中1的个数满足[s1,s2]的下一个数

分析: 从低位到高位枚举0变成1  把之后的从最低位到当前位补够1 其他的全部置0

         注意判断当前是否满足条件 不行就枚举下一个  INT_MAX下一位会炸int国际惯例开LL了

代码:

//
// Created by TaoSama on 2015-09-27
// 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 << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

long long d, s1, s2;

int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
int kase = 0;
while(t--) {
scanf("%I64d%I64d%I64d", &d, &s1, &s2);
int n = __builtin_popcountll(d);
int cnt = 0;
for(int i = 0; i < 32; ++i) {
if(d >> i & 1LL) ++cnt;
else {
long long ans = d, have = n - cnt + 1;
ans |= 1LL << i;
bool ok = false;
for(int j = 0; j < i; ++j) {
if(have < s1) {
ans |= 1LL << j;
if(++have == s1) ok = true;
} else {
ok = true;
ans &= ~(1LL << j);
}
}
int tmp = __builtin_popcountll(ans);
ok = tmp >= s1 && tmp <= s2;
if(ok) {
printf("Case #%d: %I64d\n", ++kase, ans);
break;
}
}
}
}
return 0;
}