(Problem 29)Distinct powers

Consider all integer combinations ofabfor 2≤a≤5 and 2≤b≤5:

22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125

If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

How many distinct terms are in the sequence generated byabfor 2≤a≤100 and 2≤b≤100?

题目大意:

考虑 ab 在 2 ≤ a ≤ 5,2 ≤ b ≤ 5下的所有整数组合:

22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125

如果将这些数字排序,并去除重复的,我们得到如下15个数字的序列:

4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

ab 在 2 ≤ a ≤ 100,2 ≤ b ≤ 100 下生成的序列中有多少个不同的项?

算法设计(方法1):

1、将ab 进行因数分解,以字符串的形式保存,eg.  285 = (4 * 7)5 = (22 * 7)= 2^10*7^5

2、用一个结构体数组保存所有的数的因数分解表达式

3、对上述结构体数组排序

4、遍历此数组,找出不相同的项的总数

//(Problem 29)Distinct powers
// Completed on Tue, 19 Nov 2013, 07:28
// Language: C
//
// 版权所有(C)acutus   (mail: acutus@126.com) 
// 博客地址:http://www.cnblogs.com/acutus/
#include <stdio.h> 
#include <string.h>

const int prim[25] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,41,
                      43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};

struct node
{
   char list[100];

}num[9801];

int cmp(const void *a, const void *b)
{
    return strcmp((*(struct node*)a).list, (*(struct node*)b).list);
}

char * explain(int a, int b)   /*将a^b分解因数*/
{
    char s[100], ch;
    char *p;
    p = s;
    int t;
    for(int i = 0; i < 25; i++) {
        t = 0;
        while(a % prim[i] == 0) {
            if(t == 0) {
                sprintf(p,"%d",prim[i]);
            }
            a /= prim[i];
            t++;
        }
        if(t > 0) {
            p = s + strlen(s);
            *p++ = '^';
            t = t * b;
            sprintf(p,"%d",t);
            p = s + strlen(s);
            if(a != 1) {
                *p++ = '*';
            } else {
                break;
            }
        }
    }
    return s;
}

void solve(void)
{
    int i, j, k, sum;
    k = 0;
    for(i = 2; i < 101; i++) {
        for(j = 2; j < 101; j++) {
            strcpy(num[k++].list, explain(i,j));
        }
    }
    qsort(num, 9801, sizeof(num[0]),cmp);
    sum = 1;
    for(i = 0; i < 9801; ) {
        j = i + 1;
        if(j >= 9801)  break;
        while(strcmp(num[i].list, num[j].list) == 0) {
            j++;
        }
        i = j;
        sum ++;
    }
    printf("%d\n",sum);
}

int main(void)
{
    solve();
    return 0;
}

算法设计(方法2):

仔细考察数字矩阵的规律,可以发现:

   能够发生重复的数字,将他们因数分解以后,得到的指数的底都是相同的,e.g. 16与64……,在2~100中,能够发生重复数字的底只有4、8、16、32、64、9、27、81、25、36、49、81、100,于是可以在底为2的时候就排除掉以4、8、16、32、64为底的重复的数字。

#include<stdio.h>   
#include<stdbool.h>
#include<stdlib.h>

#define N 101
#define M 601

int main(void)
{
 int answer = 0;
 int i, j, k, l;
 bool flag[M];

 bool use[N] = {false};  

 for (i = 2; i < N; i++)
 {
  if (!use[i])
  {
   int t = i;

   memset(flag, false, sizeof(flag));

   for (j = 2; j < N; j++)
   {
    t = t * i;
    if (t >= N)
    {
     break;
    }
    use[t] = true;
   }

   for (k = 1; k < j; k++)
   {
    for (l = 2; l < N; l++)
    {
     flag[k*l] = true;
    }
   }

   for (k = 2; k < M; k++)
   {
    if(flag[k]){
     answer++;
    } 
 
   }
  }
}
 printf("%d\n",answer);
 return 0;
}

 

Answer:
9183
posted @ 2014-02-12 20:59  acutus  阅读(349)  评论(0编辑  收藏  举报
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