HDU 5491 The Next （二进制） 2015合肥网络赛

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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 << endlconst 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;}

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