Question 24: Prime Number distance

The description now gives you some numbers. You need to write a program to output the adjacent prime numbers of these integers and their length. If there is a prime number with an equal-distance length between the left and right, the value on the left and the corresponding distance are output. If the input integer itself is a prime number, the output is the prime number, and the number of test data groups is N (0 <N <= 10000) from the first line of the output 0) each row of the next N Rows has an integer M (0 <M <1000000), and two integers a B are output for each row. A Indicates the prime number closest to the test data, and B indicates the distance between them. Sample input 36810 sample output 5 17 111 1 source classic question [cpp]/*********************** * *********** Date: 2013-3-26 * Author: SJF0115 * question: Question 24: Prime Number distance problem * Source: http://acm.nyist.net/JudgeOnline/problem.php? Pid = 24 * result: AC * Source: Nanyang sci-tech OJ * Summary: * *********************************/# include <stdio. h> # include <string. h> # include <math. h> # define MAX 1000010 int prime [MAX]; // Prime number table void prime () {memset (prime, 0, sizeof (prime); for (int I = 2; I <sqrt (MAX) + 1; I ++) {if (prime [I]) {continue;} else {for (int j = I * I; j <MAX; j + = I) {prime [j] = 1 ;}}} int main () {int N, M, LIndex, RIndex; // freopen ("C: \ Users \ SJF \ Users t Op \ acm.txt "," r ", stdin); Prime (); prime [1] = 1; prime [0] = 1; scanf (" % d ", & N); // N groups of test data while (N --) {scanf ("% d", & M); if (prime [M] = 0) {printf ("% d 0 \ n", M) ;}else {RIndex = M; LIndex = M; // find the prime number on the left while (prime [LIndex] = 1 & LIndex> = 0) {LIndex --;} // find the right prime number while (prime [RIndex] = 1) {RIndex ++;} // if (LIndex <0) not found on the left) {printf ("% d \ n", RIndex, RIndex-M);} // else if (M-LIndex <= RIndex-M) {Printf ("% d \ n", LIndex, M-LIndex);} else {printf ("% d \ n", RIndex, RIndex-M) ;}} return 0 ;} /********************************** Date: 2013-3-26 * Author: SJF0115 * question: Question 24: Prime Number distance problem * Source: http://acm.nyist.net/JudgeOnline/problem.php? Pid = 24 * result: AC * Source: Nanyang sci-tech OJ * Summary: * *********************************/# include <stdio. h> # include <string. h> # include <math. h> # define MAX limit 10int prime [MAX]; // Prime number table void prime () {memset (prime, 0, sizeof (prime); for (int I = 2; I <sqrt (MAX) + 1; I ++) {if (prime [I]) {continue;} else {for (int j = I * I; j <MAX; j + = I) {prime [j] = 1 ;}}} int main () {int N, M, LIndex, RIndex; // freopen ("C: \ Users \ SJF \ Desktop \ acm.txt "," r ", stdin); Prime (); prime [1] = 1; prime [0] = 1; scanf ("% d", & N); // N groups of test data while (N --) {scanf ("% d", & M ); if (prime [M] = 0) {printf ("% d 0 \ n", M);} else {RIndex = M; LIndex = M; // find the prime number on the left while (prime [LIndex] = 1 & LIndex> = 0) {LIndex --;} // find the right prime number while (prime [RIndex] = 1) {RIndex ++;} // if (LIndex <0) not found on the left) {printf ("% d \ n", RIndex, RIndex-M);} // else if (M-LIndex <= RIndex-M) {printf ("% d \ n", LIndex, M-LIndex);} else {printf ("% d \ n", RIndex, RIndex-M) ;}} return 0 ;}