Couple Number

本文探讨了如何判断一个整数是否可以表示为两个整数的平方差,并给出了一种有效的算法实现。输入两个整数n1和n2,算法会计算并返回在n1到n2范围内满足条件的整数数量。
  • Description

任何一个整数N都能表示成另外两个整数a和b的平方差吗?如果能,那么这个数N就叫做Couple number。你的工作就是判断一个数N是不是Couple number。

  • Input

仅一行,两个长整型范围内的整数n1和n2,之间用1个空格隔开。

  • Output

输出在n1到n2范围内有多少个Couple number。

注意:包括n1和n2两个数,且n1<n2,n2 - n1 <= 10000000。
  • Sample Input

1 10

  • Sample Output

7

 #include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;


int main()
{
    long n1,n2,count,k1,k2;
    cin>>n1>>n2;
    if(n1>=0||n2<=0)
    {
    if(n2<=0) {long t=n1;n1=-n2;n2=-t;}
    k2=(n2+2)/4;
    k1=(n1+2)/4;
    if((n1+2)%4)
    k1++;
    count=n2-n1+1-(k2-k1+1);
    }
    else
        count=n2-n1+1-(-n1+2)/4-(n2+2)/4;
    cout<<count<<endl;
return 0;
}

#include <stdio.h> #include <stdlib.h> #include <string.h> typedef struct{ long int male; long int female; }couple; int main(void){ int couple_num; scanf("%d",&couple_num); getchar(); couple couple_set[couple_num]; for(int n = 0;n<couple_num;n++){ char couple_id[12]; fgets(couple_id,12,stdin); char *endptr; char * token = strtok(couple_id," \n"); long int num1 = strtol(token,&endptr,10); token = strtok(NULL," \n"); long int num2 = strtol(token,&endptr,10); (couple_set+n)->male = num1; (couple_set+n)->female = num2; /*while(token != NULL){ int order = 1; long int num = strtol(token,10);//接应用出问题 (couple_set[n]).f = num; token = strtok(NULL," \n"); order++; }*/ } int arrive_people; scanf("%d",&arrive_people); getchar(); char people_arrive_str[5*arrive_people+1]; fgets(people_arrive_str,5*arrive_people+1,stdin); int locate_people[arrive_people]; char * token2 = strtok(people_arrive_str," \n"); int mm = 0; while(token2 != NULL){ char *endptr; long int num2 = strtol(token2,&endptr,10); locate_people[mm]= num2; mm++; token2 = strtok(NULL," \n");} int gouliangvalue[arrive_people]; for(int q = 0;q<arrive_people;q++){ gouliangvalue[q] = 0; } for(int k = 0;k<couple_num;k++){ int count = 0; int left,right; for(int p = 0;p<arrive_people;p++){ if(locate_people[p]==couple_set[k].male){count++;left = p;} if(locate_people[p]==couple_set[k].female){count++;right = p;} }if(count==2){ gouliangvalue[left] = -49999; gouliangvalue[right] = -49999; if(left>right){ int temp = left; left = right; right = temp; } int distance = (right - left); if(distance==1){gouliangvalue[left-1]+=1;gouliangvalue[right+1]+=1;} if(distance!=1){for(int m = left+1;m<right;m++){ gouliangvalue[m]+=1; } }}for(int h = 0;h<arrive_people;h++){ int max = gouliangvalue[0]; int index = 0; if(gouliangvalue[h]>max){max = gouliangvalue[h];index = h;} printf("%d ",index); }} return 0;}
09-25
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