USACO 2.3.3 Zero Sum 解题报告

Zero Sum

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)

7

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

题目意思很简单个，就是在序列1…N中每两个连续的数字间指定一个操作符（+、-或空格），使得最后的结果为0 输出这样的结果（有多个就按ASCII 码的顺序输出）

这题可以用DFS直接求解，通过在每两个连续的数间指定可能的运算符（按空格，+，-的顺序指定）。当运算到最后一个数时如果结果为0，则输出

参考代码

/* 

ID:shiryuw1 

PROG:zerosum 

LANG:C++ 

*/ 

#include<iostream> 

#include<cstdlib> 

#include<cstdio> 

#include<cstring> 

#include<algorithm> 

#include<cmath> 

using namespace std; 

int n; 

struct prog{ 

    char c[10]; 

    int pre; 

    int next; 

    char op; 

    int k; 

}; 

int add (int a,int b,char c) 

{ 

    switch(c) 

    { 

    case 0:return a;break; 

    case '+':return a+b;break; 

    case '-':return a-b;break; 

    } 

} 

void DFS(prog ans) 

{ 

    int i; 

    if(ans.k==n) 

    { 

        if(add(ans.pre,ans.next,ans.op)==0) 

        { 

            for(i=1;i<=n;i++) 

            { 

                if(i!=1) 

                    cout<<ans.c[i]; 

                cout<<i; 

            } 

            cout<<endl; 

        } 

        return ; 

    } 

    ans.k++; 

    prog tmp; 

    if(ans.op==0) 

    { 

        tmp=ans; 

        tmp.c[tmp.k]=' '; 

        tmp.pre=tmp.pre*10+tmp.k; 

        DFS(tmp);





        tmp=ans; 

        tmp.c[tmp.k]='+'; 

        tmp.next=tmp.k; 

        tmp.op='+'; 

        DFS(tmp); 

        tmp=ans; 

        tmp.c[tmp.k]='-'; 

        tmp.next=tmp.k; 

        tmp.op='-'; 

        DFS(tmp); 

    } 

    else 

    { 

        tmp=ans; 

        tmp.c[tmp.k]=' '; 

        tmp.next=tmp.next*10+tmp.k; 

        DFS(tmp);



        tmp=ans; 

        tmp.c[tmp.k]='+'; 

        tmp.pre=add(tmp.pre,tmp.next,tmp.op); 

        tmp.next=tmp.k; 

        tmp.op='+'; 

        DFS(tmp);



        tmp=ans; 

        tmp.c[tmp.k]='-'; 

        tmp.pre=add(tmp.pre,tmp.next,tmp.op); 

        tmp.next=ans.k; 

        tmp.op='-'; 

        DFS(tmp); 

    } 

} 

int main() 

{ 

    freopen("zerosum.in","r",stdin); 

    freopen("zerosum.out","w",stdout); 

    cin>>n; 

    prog ans; 

    ans.k=1; 

    ans.pre=1; 

    ans.op=0; 

    ans.next=0; 

    ans.c[0]=0; 

    DFS(ans); 

    return 0; 

}