2015 Baidu STAR Program Design Competition, 2015 STAR Program Design

2015 Baidu STAR Program Design Competition, 2015 STAR Program Design

[Question link ]:Cilck here ~~

[General idea ]:

In today's world, the network is everywhere. Due to a mistake made by Xiao duxiong, as a network administrator of du, he has a large number of IP addresses on his hand, du Xiong wants to know the number of network addresses under a fixed subnet mask. The result of bitwise calculation of the subnet mask and IP address. For example:

Subnet Mask: A. B .C.D

IP Address: a. B. c. d

Network Address: (A & a). (B & B). (C & c). (D & d)

This is a simple question. You can sort and process it. Because the bool cmp function is wrong, it is still written too little after one afternoon of debugging ~~

Code:

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;const int N=1e5+5;struct node{    int a1,a2,a3,a4;} ip[N],ans[N],res[N];bool cmp(node a,node b){    if(a.a1!=b.a1) return a.a1<=b.a1;    if(a.a2!=b.a2) return a.a2<=b.a2;    if(a.a3!=b.a3) return a.a3<=b.a3;    if(a.a4!=b.a4) return a.a4<=b.a4;}int main(){    int t,tot=1,n,m;    cin>>t;    while(t--)    {        memset(res,0,sizeof(res));        cin>>n>>m;        for(int i=1; i<=n; i++) scanf("%d.%d.%d.%d",&ip[i].a1,&ip[i].a2,&ip[i].a3,&ip[i].a4);        for(int i=1; i<=m; i++) scanf("%d.%d.%d.%d",&ans[i].a1,&ans[i].a2,&ans[i].a3,&ans[i].a4);        int k=0;        printf("Case #%d:\n",tot++);        for(int i=1; i<=m; i++)        {            int c=0,cnt=0;            for(int j=1; j<=n; j++)            {                res[k].a1=ip[j].a1&ans[i].a1;                res[k].a2=ip[j].a2&ans[i].a2;                res[k].a3=ip[j].a3&ans[i].a3;                res[k].a4=ip[j].a4&ans[i].a4;                k++;            }            sort(res+k-n,res+k,cmp);            for(int j=k-n; j<k; j++)            {                if((res[j].a1==res[j+1].a1)&&(res[j].a2==res[j+1].a2)&&(res[j].a3==res[j+1].a3)&&(res[j].a4==res[j+1].a4))                    c++;            }            printf("%d\n",n-c);        }    }    return 0;}