如何找到在C n次贷 [英] How to find the nth prime in C

问题描述

这个应用程序将收到一个号码 N 。接收到该号码后，该方案具有显示 N -th中素数的列表中素数。例如，如果用户输入 3 ，程序应该显示 5 ，因为 5 是第三个首相起点在 2 。我知道什么是错我的code，但我不知道问题出在哪里，我怎么能解决这个问题。

的#include＆LT;＆stdio.h中GT;诠释主（）{
    诠释N，I，旗，素;
    INT计数器= 1;
    scanf函数（％d个，＆安培; N）;
    如果（N == 1）黄金= 2;
    其他
        做{
            素= 3;
            对于（i = 2; I＆LT;素;我++）{
                标志= 1;
                如果（素％我== 0）{
                    标志= 0;
                }
            }
            如果（标志== 1）
                反++;
            素++;
        }而（计数器= N！）;
    如果（反== N）
    的printf（％d个\\ N，素数）;
    返回0;
}


解决方案

你这样的政策的遗体修复示例：

 ＃包括LT＆;＆stdio.h中GT;诠释主要（无效）{
    诠释N，I，旗，素;
    INT计数器= 1;    scanf函数（％d个，＆安培; N）;
    如果（N == 1）
        素= 2;
    其他{
        素= 1;
        做{
            黄金+ = 2;
            标志= 1;
            对于（i = 3; I＆LT;素，I + = 2）{
                如果（素％我== 0）{
                    标志= 0;
                    打破;
                }
            }
            如果（标志== 1）
                反++;
        }而（计数器= N！）;
    }
    的printf（％d个\\ N，素数）;    返回0;
}
 

This application will receive a number n. After receiving this number, the program has to show the n-th prime in the list of primes. For example, if the user enters 3, the program is supposed to display 5, because 5 is the third prime starting at 2. I know that something is wrong with my code but I don't know where the problem is and how I can fix it.

#include <stdio.h>

int main() {
    int n, i, flag, prime;
    int counter = 1;
    scanf("%d", &n);
    if (n == 1) prime = 2;
    else
        do{
            prime = 3;
            for (i = 2; i < prime; i++) {
                flag = 1;
                if (prime % i == 0) {
                    flag = 0;
                }
            }
            if (flag == 1)
                counter++;
            prime++;
        } while (counter != n);
    if (counter == n)
    printf("%d\n", prime);
    return 0;
}

解决方案

Fix sample of remains of your policy like this :

#include<stdio.h>

int main(void){
    int n, i, flag, prime;
    int counter = 1;

    scanf("%d", &n);
    if (n == 1)
        prime = 2;
    else {
        prime = 1;
        do{
            prime += 2;
            flag = 1;
            for (i = 3; i < prime; i+=2){
                if (prime % i == 0) {
                    flag = 0;
                    break;
                }
            }
            if(flag == 1)
                counter++;
        } while (counter != n);
    }
    printf("%d\n", prime);

    return 0;
}

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