UVA725 Division【枚举】

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  Division 

Each line of the input file consists of a valid integer N. An input of zero is to terminate the program.

Your program have to display ALL qualifying pairs of numerals, sorted by increasing numerator (and, of course, denominator).

0

94736 / 01528 = 62

题目大意：输入一个正整数N，要求从小到大的输出形如abcde / fghij = n的表达式。

要求a~f为数字0~9，且不能重复(前边可有0)。

思路：如果将a b c d e f g h i j全部遍历的话，复杂度是10!，没有必要，直接枚举

f g h i j，然后算出a b c d e，去判断是否存在重复数字即可。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

int main()
{
    int sum,num,N,j;
    int a,b,c,d,e;
    int f,g,h,l,m;
    int x[10],ok = 0;
    while(~scanf("%d",&N) && N)
    {
        if(ok == 0)
            ok = 1;
        else
            printf("\n");
        sum = 0;
        for(int i = 1234; i <= 98765; i++)
        {
            memset(x,0,sizeof(x));
            a = i%10,b = i/10%10,c = i/100%10,d = i/1000%10,e = i/10000%10;
            num = i*N;
            if(num>98765)
                break;
            f = num%10,g = num/10%10,h = num/100%10,l = num/1000%10,m = num/10000%10;
            x[a]++,x[b]++,x[c]++,x[d]++,x[e]++;
            x[f]++,x[g]++,x[h]++,x[l]++,x[m]++;
            for(j = 0; j <= 9; j++)
                if(x[j]>1)
                    break;
            if(j==10)
            {
                printf("%d%d%d%d%d / %d%d%d%d%d = %d\n",m,l,h,g,f,e,d,c,b,a,N);
                sum++;
            }
            else
                continue;
        }
        if(sum==0)
            printf("There are no solutions for %d.\n",N);
    }

    return 0;
}

UVA725 Division【枚举】

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原文地址：http://blog.csdn.net/lianai911/article/details/41939879