Prime permutations
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.
There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
What 12-digit number do you form by concatenating the three terms in this sequence?
素数重排
公差为3330的三项等差序列1487、4817、8147在两个方面非常特别:其一,每一项都是素数;其二,两两都是重新排列的关系。
一位素数、两位素数和三位素数都无法构成满足这些性质的数列,但存在另一个由四位素数构成的递增序列也满足这些性质。
将这个数列的三项连接起来得到的12位数是多少?
//(Problem 49)Prime permutations
// Completed on Thu, 13 Feb 2014, 15:35
// Language: C
//**********************************************
// 版权所有(C)wu yudong
// 博客地址:http://www.wuyudong.com/
//**********************************************
#include<stdio.h>
#include<stdbool.h>
#include<stdlib.h>
#include<string.h>
int a[1230];
bool prim(int n)
{
int i;
for(i = 2; i * i <= n; i++) {
if(n % i ==0) return false;
}
return true;
}
int cmp(const void *a, const void *b)
{
return (*(char*)a - *(char*)b);
}
void init()
{
int i, j;
i = 3;
j = 1;
a[0] = 2;
while(j < 1230) {
if(prim(i)) {
a[j++] = i;
}
i += 2;
}
}
bool judge(int a, int b, int c)
{
char A[5], B[5], C[5];
sprintf(A, "%d", a);
qsort(A, 4, sizeof(char), cmp);
sprintf(B, "%d", b);
qsort(B, 4, sizeof(char), cmp);
sprintf(C, "%d", c);
qsort(C, 4, sizeof(char), cmp);
if(strcmp(A, B)== 0 && strcmp(A, C) == 0)
return true;
return false;
}
void solve()
{
int i, b, c, d;
i = 0;
init();
while(a[i++] < 1000);
for(; i < 1229; i++) {
b = a[i]; c = a[i] + 3330; d = a[i] + 6660;
if(d < 9999) {
if(prim(b) && prim(c) && prim(d)) {
if(judge(b, c, d)) {
printf("%d %d %d\n", b, c, d);
}
}
}
}
}
int main()
{
solve();
return 0;
}
Answer:296962999629
Completed on Thu, 13 Feb 2014, 15:35
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