Gym

题意：给定一个数字，让你构造成一些表达式，最后结果是该数字的概率要大于50%。

析：我们可以把一个数分解是2的多少次幂，然后加起来就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 500 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
map<int,char> ID;
int f[20];

void init(){
  f[1] = 1;
  for(int i = 2; i <= 15; ++i)  f[i] = 2 * f[i-1];
  ID[1] = 'a';  ID[2] = 'b';  ID[4] = 'c';
  ID[8] = 'd';  ID[16] = 'e'; ID[32] = 'f';
  ID[64] = 'g'; ID[128] = 'h';
  printf("%c = ? max ?
", 9 + 'a');
  for(int i = 10; i < 26; ++i)  printf("%c = (%c max %c)
", i+'a', i-1+'a', i-1+'a');
  printf("a = (z max z) / z
");
  for(int i = 2; i <= 128; i *= 2)
    printf("%c = %c + %c
",ID[i], ID[i>>1],ID[i>>1]);
}

int main(){
  freopen("java2016.in","r",stdin);
  freopen("java2016.out","w",stdout);
  while(scanf("%d", &n) == 1){
    if(n == 0){ printf("? /?/ ?
");  continue; }
    init();
    bool flag = false;
    for(int i = 8; i >= 1; --i){
      if(n >= f[i]){
        if(flag)  printf(" + ");
        flag = true;
        printf("%c", ID[f[i]]);
        n -= f[i];
      }
    }
    printf("
",'
');
  }
  return 0;
}