我想通过将素数放入一个子集来组织一组整数.
例如:
输入集:
{} 2,4,6,7,11,6,3.
期望的结果:
素数:{2,7,11,3}.
非素数:{4,6,6}.
这是我合理的代码,结果是一个荒谬的结果:
#include <stdio.h>
int main() {
int i,j,n,a=0,b1=0,b2=0;
printf(" Enter the number of elements in the set: ");
scanf("%d",&n);
int integer[n],nonPrime[n],prime[n];
for(i=1;i<=n;i++) {
printf(" Enter #%d element of the set: ",i);
scanf("%d",&integer[i-1]);
}
printf("\n The set is: {");
for(i=1;i<=n;i++) {
printf("%d ",integer[i-1]);
}
printf("}");
for(i=1;i<=n;i++) {
for(j=2;j<=integer[i-1]/2;j++) {
if(integer[i-1]%j==0) {
a=1;
break;
}
}
if(a==1) {
integer[i-1]=nonPrime[i-1];
b1++;
}
else {
integer[i-1]=prime[i-1];
b2++;
}
}
printf("\n Prime numbers: { ");
for(i=1;i<=b2;i++) {
printf("%d ",prime[i-1]);
}
printf("}\n Non prime numbers: { ");
for(i=1;i<=b1;i++) {
printf("%d ",nonPrime[i-1]);
}
printf("} \n");
return 0;
}
//HENG SOK MENG
输出:
Enter the number of elements in the set: 6
Enter #1 element of the set: 1
Enter #2 element of the set: 3
Enter #3 element of the set: 5
Enter #4 element of the set: 7
Enter #5 element of the set: 4
Enter #6 element of the set: 6
The set is: {1 3 5 7 4 6 }
Prime numbers: { 1965421290 1965972381 718360966 32 }
Non prime numbers: { 2686560 718361022 }
Process returned 0 (0x0) execution time : 6.063 s
Press any key to continue.
最佳答案 我认为你的逻辑有些错误.此代码显示您的循环可以简化为(i = 0; i< n; i),这将允许索引[i]而不是[i-1]. 您可以使用像isprime(int n)这样的辅助函数来检查整数[n]数组中的素数,然后相应地添加到其他相应的分区数组. 此代码段演示了这一点:
#include <stdio.h>
#include <stdlib.h>
int isprime(int n);
int
main(void) {
int n, i, primecnt = 0, nonprimecnt = 0;
printf("Enter the number of elements in the set: ");
if (scanf("%d", &n) != 1) {
printf("Invalid entry\n");
exit(EXIT_FAILURE);
}
int integers[n], prime[n], nonprime[n];
for (i = 0; i < n; i++) {
printf(" Enter #%d element of the set: ",i+1);
if (scanf("%d", &integers[i]) != 1) {
printf("Invalid entry\n");
exit(EXIT_FAILURE);
}
}
printf("\n The set is: {");
for(i = 0;i < n; i++) {
printf("%d ",integers[i]);
}
printf("}");
for (i = 0; i < n; i++) {
if (isprime(integers[i])) {
prime[primecnt++] = integers[i];
} else {
nonprime[nonprimecnt++] = integers[i];
}
}
printf("\n Prime numbers: { ");
for(i = 0; i < primecnt; i++) {
printf("%d ", prime[i]);
}
printf("}\n Non prime numbers: { ");
for(i = 0; i < nonprimecnt; i++) {
printf("%d ",nonprime[i]);
}
printf("} \n");
return 0;
}
int
isprime(int n) {
int divisor;
if (n < 2) {
return 0;
}
for (divisor=2; divisor*divisor<=n; divisor++) {
if (n%divisor==0) {
return 0;
}
}
return 1;
}