Day 6 jzoj 4314. 【NOIP2015模拟11.4】老司机

Description

Input

Output

Sample Input

4
1 3 6 6

Sample Output

3
1 2 3

Data Constraint

做法：居然暴力就可以过233333
代码如下：

#include <cstdio>
#include <iostream>
#include <cstring>
#define rep(i, a, b)    for (int i = a; i <= b; i++)
using namespace std;
int n, a[30], b[30], tm;
bool bobo;

bool judge(int x, int mdep, int dep, int sum)
{
    if (x == sum)   return 1;
    if (x < sum)    return 0;
    if (dep > mdep) return 0;
    if (judge(x, mdep, dep + 1, sum + b[dep]))  return 1;
    return judge(x, mdep, dep + 1, sum);
}

void print(int x)
{
    bobo = 1;
    printf("%d\n", x);
    rep(i, 1, x)
        printf("%d ", b[i]);
    return; 
}

void dfs(int dep, int last, int mdep)
{
    if (bobo)   return;
    if (dep > mdep)
    {
        rep(i, 1, n)
            if (a[i] != a[i - 1] && !judge(a[i], mdep, 1, 0))   return;
        print(mdep);
        return;
    }
    rep(i, last, tm)
    {
        b[dep] = i;
        dfs(dep + 1, i, mdep);
        b[dep] = 0;
    }
}

int main()
{
    freopen("driver.in", "r", stdin);
    freopen("driver.out", "w", stdout);
    scanf("%d", &n);
    rep(i, 1, n)
    {
        scanf("%d", &a[i]);
        tm = max(tm, a[i]);
    }
    rep(i, 1, n)    if(!bobo)   dfs(1, 1, i);
}