# Binary Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 42    Accepted Submission(s): 25
Special Judge

Problem Description The Old Frog King lives on the root of an infinite tree. According to the law, each node should connect to exactly two nodes on the next level, forming a full binary tree.

Since the king is professional in math, he sets a number to each node. Specifically, the root of the tree, where the King lives, is   1. Say   froot=1.

And for each node   u, labels as   fu, the left child is   fu×2  and right child is   fu×2+1. The king looks at his tree kingdom, and feels satisfied.

Time flies, and the frog king gets sick. According to the old dark magic, there is a way for the king to live for another   N  years, only if he could collect exactly   Nsoul gems.

Initially the king has zero soul gems, and he is now at the root. He will walk down, choosing left or right child to continue. Each time at node   x, the number at the node is   fx  (remember   froot=1), he can choose to increase his number of soul gem by   fx, or decrease it by   fx.

He will walk from the root, visit exactly   K  nodes (including the root), and do the increasement or decreasement as told. If at last the number is   N, then he will succeed.

Noting as the soul gem is some kind of magic, the number of soul gems the king has could be negative.

Given   N,   K, help the King find a way to collect exactly   N  soul gems by visiting exactly   K  nodes.
Input First line contains an integer   T, which indicates the number of test cases.

Every test case contains two integers   N  and   K, which indicates soul gems the frog king want to collect and number of nodes he can visit.

1T100.

1N109.

N2K260.
Output For every test case, you should output ” Case #x:” first, where   x  indicates the case number and counts from   1.

Then   K  lines follows, each line is formated as ‘a b’, where   a  is node label of the node the frog visited, and   b  is either ‘+’ or ‘-‘ which means he increases / decreases his number by   a.

It’s guaranteed that there are at least one solution and if there are more than one solutions, you can output any of them.

Sample Input

2 5 3 10 4
Sample Output

Case #1: 1 + 3 – 7 + Case #2: 1 + 3 + 6 – 12 +
Source 2015ACM/ICPC亚洲区上海站-重现赛（感谢华东理工）

Sample Input

2 5 3 10 4

Sample Output

Case #1: 1 + 3 – 7 + Case #2: 1 + 3 + 6 – 12 +

#include<stdio.h>
#include<string.h>
int casenum,casei;
int n,m;
long long b[64],s[64],num[64];
int main()
{
for(int i=1;i<=62;++i)b[i]=1ll<<i-1;
for(int i=0;i<=62;++i)s[i]=(1ll<<i)-1;
scanf("%d",&casenum);
for(casei=1;casei<=casenum;++casei)
{
scanf("%d%d",&n,&m);
printf("Case #%d:\n",casei);
memcpy(num,b,sizeof(b));
long long dif=s[m]-n;
if(dif&1){++dif;++num[m];}
dif>>=1;
for(int i=1;i<=m;++i)printf("%lld %c\n",num[i],dif&b[i]?'-':'+');
}
return 0;
}

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=0,M=0,Z=1e9+7,ms63=1061109567;
int casenum,casei;
int n,m;
LL b[64],s[64],num[64];
int main()
{
for(int i=1;i<=62;++i)b[i]=1ll<<i-1;
for(int i=0;i<=62;++i)s[i]=(1ll<<i)-1;
scanf("%d",&casenum);
for(casei=1;casei<=casenum;++casei)
{
scanf("%d%d",&n,&m);
printf("Case #%d:\n",casei);
MC(num,b);
LL dif=s[m]-n;
if(dif&1){++dif;++num[m];}
dif>>=1;
for(int i=1;i<=m;++i)printf("%lld %c\n",num[i],dif&b[i]?'-':'+');
}
return 0;
}
/*
【trick&&吐槽】

【题意】

【类型】

【分析】

【时间复杂度&&优化】
O(Tm)

*/

原文作者：B树
原文地址: https://blog.csdn.net/snowy_smile/article/details/50094987
本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。