PAT 1031B 数素数
令 Pi 表示第 i 个素数。现任给两个正整数 M≤N≤104,请输出 PM到 PN的所有素数。
输入格式:
输入在一行中给出 M 和 N,其间以空格分隔。
输出格式:
输出从 PM到 PN的所有素数,每 10 个数字占 1 行,其间以空格分隔,但行末不得有多余空格。
输入样例:
5 27
输出样例:
11 13 17 19 23 29 31 37 41 43
47 53 59 61 67 71 73 79 83 89
97 101 103
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
//判断是否为素数
bool isPrime(int number){
for(int i = 2;i<=sqrt(number);i++){
if(number%i == 0){
return false;
}
}
return true;
}
int main(){
int m,n,k,number;
cin>>m>>n;
k = 0;
number = 2;
while(k<=n){
if(isPrime(number)){
k++;
if(k>=m && k<=n){
if((k-m+1)%10 != 1){
cout<<" ";
}
cout<<number;
if((k-m+1)%10 == 0){
cout<<endl;
}
}
}
number++;
}
return 0;
}
PAT1059A Prime Factors
Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pm km
Input Specification:
Each input file contains one test case which gives a positive integer N in the range of long int.
Output Specification:
Factor N in the format N = N = p1 k1*p2k2*…pm^km
, where p i’s are prime factors of N in increasing order, and the exponent kiis the number of pi-- hence when there is only on pi, ki is 1 and must NOT be printed out.
Sample Input:
97532468
Sample Output:
97532468=2^21117101*1291
#include<cstdio>
#include<iostream>
#include<cmath>
/*
*生成素数数组
*得出素数表
* 对输入内容进行取余
*/
using namespace std;
const int maxn = 100010;
//判断是否为素数
bool isPrime(int n){
if(n == 1) return false;
int rs = (int)sqrt(n*1.0);
for(int i = 2;i<=rs;i++){
if(n%i == 0){
return false;
}
}
return true;
}
int prime[maxn],pNum=0;
//求素数表
void getPrime(){
for(int n =1;n<maxn;n++){
if(isPrime(n) == true){
prime[pNum++] = n;
}
}
}
//创建结构体,存放素数和该素数的个数
struct factor{
int x,cnt;
}fac[10];
int main(){
getPrime();
int n,num=0;
scanf("%d",&n);
if (n==1) printf("1=1");
else{
printf("%d=",n);
int sqr = (int)sqrt(1.0*n);
for(int i=0;i<pNum&&prime[i]<=sqr;i++){
if(n%prime[i]==0){
fac[num].x = prime[i];
fac[num].cnt = 0;
while(n%prime[i]==0){
fac[num].cnt++;
n/=prime[i];
}
num++;
}
if (n==1) break;
}
if (n!=1){
fac[num].x = n;
fac[num++].cnt=1;
}
for(int i=0;i<num;i++){
if(i>0) printf("*");
printf("%d",fac[i].x);
if(fac[i].cnt>1){
printf("^%d",fac[i].cnt);
}
}
}
return 0;
}
大整数尝试
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
//大整数,高位存在数组高位
//创建大整数结构体
struct gibn{
int d[1000];
int len;
bign{
memset(d, 0, sizeof(d));
len = 0;
}
};
//创建大整数结构体变量
bign change(char str[]){
bign a;
a.len = strlen(str);
for(int i =0;i<a.len;i++){
a.d[i] = str[a.len-1-i]-'0';
}
return a;
}
//对两个大整数进行比较
int compare(bign a,bign b){
if(a.len>b.len)
return 1;
else if(a.len < b.len)
return -1;
else{
for(int i =a.len-1;i>=0;i--){
if(a.d[i]>b.d[i])
return 1;
else if(a.d[i]<b.d[i])
return -1;
}
return 0;
}
}
//进行加法
//取余10得出的数据保存在当前位,除10得出的数据存到下一位
gibn add(gibn a,gibn b){
bign c;
int carry = 0;
for(int i =0;i<a.len||i<b.len;i++){
int temp = a[i]+b[i]+carry;
c.d[c.len++] = temp%10;
carry = temp/10;
}
if(carry != 0){
c.d[c.len++] = carry;
}
return c;
}
//进行减法
gibn sub(gibn a,gibn b){
bign c;
for(int i =0;i<a.len||i<b.len;i++){
//如果当前位不够减
if(a.d[i]<b.d[i]){
a.d[i+1]--;
a.d[i] += 10;
}
c.d[c.len++] = a.d[i] - b.d[i];
}
while(c.len-1>=1 && c.d[c.len-1] == 0){
c.len--;
}
return c;
}
//进行乘法
gibn multi(gibn a, gibn b) {
bign c;
int carry = 0;
for(int i =0;i<a.len||i<b.len;i++){
int temp = a[i]*b[i]+carry;
c.d[c.len++] = temp%10;
carry = temp/10;
}
while(carry != 0){
c.d[c.len++] = carry%10;
carry = carry/10;
}
return c;
}
int main(){
char str1[1000],str2[1000];
cin>>str1;
cin>>str2;
bign a = change(str1);
bign b = change(str2);
return o;
}