【其他】【USACO】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+71+2-3-4+5+6-71-2 3+4+5+6+71-2 3-4 5+6 71-2+3+4-5+6-71-2-3-4-5+6+7

描述

请考虑一个由1到N（N=3, 4, 5 ... 9）的数字组成的递增数列：1 2 3 ... N。现在请在数列中插入“+”表示加，或者“-”表示减，“ ”表示空白(例如1-2 3就等于1-23)，来将每一对数字组合在一起（请不在第一个数字前插入符号）。计算该表达式的结果并判断其值是否为0。请你写一个程序找出所有产生和为零的长度为N的数列。

格式

PROGRAM NAME: zerosum

INPUT FORMAT

单独的一行表示整数N (3 <= N <= 9)。

OUTPUT FORMAT

按照ASCII码的顺序，输出所有在每对数字间插入“+”, “-”, 或 “ ”后能得到和为零的数列。

SAMPLE INPUT (file zerosum.in)

7

 

SAMPLE OUTPUT (file zerosum.out)

1+2-3+4-5-6+71+2-3-4+5+6-71-2 3+4+5+6+71-2 3-4 5+6 71-2+3+4-5+6-71-2-3-4-5+6+7

来自"http://www.nocow.cn/index.php/Translate:USACO/zerosum"

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模拟 枚举所有位置的符号，然后判断是否满足条件

 

/*ID:sephiro5LANG:C++PROG:zerosum*/#include<stdio.h>const char t[3]={' ','+','-'};int ans,n,len;char tes[100];void outout(char *s){ for (int i=1;i<len;++i) printf("%c",s[i]); printf("%c/n",s[len]); return;}void check(char *s){ int te=0,teans=0; bool plus=true; for (int i=1;i<=len;++i) if (tes[i]>='0'&&tes[i]<='9') te=te*10+tes[i]-'0'; else if (tes[i]=='+') { if (plus) { teans+=te; te=0; } if (!plus) { teans-=te; te=0; plus=true; } }else if (tes[i]=='-') { if (plus) { teans+=te; te=0; plus=false; } if (!plus) { teans-=te; te=0; } } if (plus) teans+=te; else teans-=te; if (teans==0) outout(tes); return ;}void make(int x){ if (x==n) { check(tes); return ; } for (int i=0;i<3;++i) { tes[++len]=t[i]; tes[++len]='0'+x+1; make(x+1); len-=2; }}int main(){ freopen("zerosum.in","r",stdin); freopen("zerosum.out","w",stdout); scanf("%d",&n); len=1; tes[1]='1'; make(1); return 0;}