标签：-o 情况 false return break represent tput bre set
InputThe first line of input contains a number TT indicating the number of test cases (T≤300000T≤300000).
Each test case consists of three integers DD, S1S1, and S2S2, as described above. It is guaranteed that 0≤D<2310≤D<231 and DD is a WYH number.
OutputFor each test case, output a single line consisting of “Case #X: Y”. XX is the test case number starting from 1. YY is the next WYH number.Sample Input
3 11 2 4 22 3 3 15 2 5
Sample Output
Case #1: 12 Case #2: 25 Case #3: 17
题目需要求的是比d大的且转化为二进制后1的个数在s1和s2之间的最小的数
开始想的是从d开始判断yi的个数分比s1小，在s1、s2之间（这里考虑的特别复杂），比s2大三种情况考虑，结果写了一大堆判断最后完美wa
后来在网上看别人的代码，是先让d+1，因为最后得到的数是比d大的，然后也是三种情况考虑，但是如果在s1、s2之间就可以直接输出，接着是比s1小则遇到0直接变成1，比s2大则遇到1变成0然后再一次进位
#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<string> #include<cmath> #include<queue> #define debug(a) cout << #a << " " << a << endl using namespace std; typedef long long ll; int a[50], j; ll sum() { ll ans = 0, t = 1; for( int i = 0; i <= 33; i ++ ) { ans = ans + a[i] * t; t *= 2; } return ans; } int main() { std::ios::sync_with_stdio(false); int T, cnt = 0; cin >> T; while( T -- ) { cnt ++; ll n, num = 0, x, y; j = 0; cin >> n >> x >> y; n ++; memset( a, 0, sizeof(a) ); while( n ) { if( n % 2 == 1 ) { a[j++] = 1; num ++; } else { a[j++] = 0; } n /= 2; } while( 1 ) { if( num >= x && num <= y ) { cout << "Case #" << cnt << ": " << sum() << endl; break; } if( num < x ) { for( int i = 0; ; i ++ ) { if( a[i] == 0 ) { a[i] = 1; num ++; break; } } } else { int i = 0; while( a[i] == 0 ) { i ++; } a[i] ++; while( a[i] == 2 ) { a[i] = 0; num --; a[i+1] ++; i ++; } num ++; } } } return 0; }
标签：-o 情况 false return break represent tput bre set
原文地址：https://www.cnblogs.com/l609929321/p/9013054.html
