USACO 2.3 Zero Sum 题解

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Zero Sum

DESCRIPTION

    Consider the sequence of digits from 1 through N (where N=9) in increasing order: 1 2 3 ... N. 
    Now insert either a `+' for addition or a `-' for subtraction or a ` ' [blank] to run the digits together between each pair of digits (not in front of the first digit). Calculate the result that of the expression and see if you get zero. 
    Write a program that will find all sequences of length N that produce a zero sum.

PROGRAM NAME: zerosum

INPUT FORMAT

A single line with the integer N (3 <= N <= 9).

SAMPLE INPUT (file zerosum.in)

OUTPUT FORMAT

In ASCII order, show each sequence that can create 0 sum with a `+', `-', or ` ' between each pair of numbers.

SAMPLE OUTPUT (file zerosum.out)

1+2-3+4-5-6+7
1+2-3-4+5+6-7
1-2 3+4+5+6+7
1-2 3-4 5+6 7
1-2+3+4-5+6-7
1-2-3-4-5+6+7

THOUGHTS:

注意n<=9，对于如此小的数据范围，我们第一反应便是暴力(DFS)，复杂度3^n，完全可以通过。
所以这道题非常简单，只需要Dfs到最后判定一下是否式子的结果为0即可。
至于字典序输出，我们观察样例，可以发现’ ’ < ‘+’ < ‘-‘，那么只需要分别让012代表’ ‘, ‘+’, ‘-‘即可。

DIFFICULTIES: pj-

AC PROGRAM:

/*
ID: kongse_1
PROG: zerosum
LANG: C++
*/

// Skq_Liao

#include <bits/stdc++.h>
using namespace std;

#define FOR(i, a, b) for (register int i = (a), i##_end_ = (b); i < i##_end_; ++i)
#define ROF(i, a, b) for (register int i = (a), i##_end_ = (b); i > i##_end_; --i)
#define debug(...) fprintf(stderr, __VA_ARGS__)

const char Fin[] = "zerosum.in";
const char Fout[] = "zerosum.out";

void In()
{
    freopen(Fin, "r", stdin);
    freopen(Fout, "w", stdout);
    return ;
}

const int MAXN = 15;

int n;
int A[MAXN];
int B[MAXN];

void Print()
{
    FOR(i, 1, n)
    {
        printf("%d", i);
        switch (A[i])
        {
            case 0:
                printf(" ");
                break;
            case 1:
                printf("+");
                break;
            default:
                printf("-");
        }
    }
    printf("%d\n", n);
    return ;
}

void Check()
{
    int last(1), num(0), ans(0);
    memset(B, 0, sizeof B);
    A[n] = 1;
    FOR(i, 1, n + 1)
    {
        B[num] = B[num] * 10 + i;
        if(A[i])
        {
            if(last == 2)
                B[num] = -B[num];
            last = A[i];
            ++num;
        }   
    }
    FOR(i, 0, num)
        ans += B[i];
    if(!ans)
        Print();
    return ;
}

void Dfs(int cur)
{
    if(cur == n)
        Check();
    else
    {
        FOR(i, 0, 3)
        {
            A[cur] = i;
            Dfs(cur + 1);
        }
    }
    return ;
}

int main()
{
    In();
    scanf("%d", &n);
    Dfs(1);
    return 0;
}