1055 The World's Richest
Forbes magazine publishes every year its list of billionaires based on the annual ranking of the world‘s wealthiest people. Now you are supposed to simulate this job,but concentrate only on the people in a certain range of ages. That is,given the net worths of?N?people,you must find the?M?richest people in a given range of their ages. Input Specification:Each input file contains one test case. For each case,the first line contains 2 positive integers:?N?(≤) - the total number of people,and?K?(≤) - the number of queries. Then?N?lines follow,each contains the name (string of no more than 8 characters without space),age (integer in (0,200]),and the net worth (integer in [?]) of a person. Finally there are?K?lines of queries,each contains three positive integers:?M?(≤) - the maximum number of outputs,and [ Output Specification:For each query,first print in a line? Name Age Net_Worth The outputs must be in non-increasing order of the net worths. In case there are equal worths,it must be in non-decreasing order of the ages. If both worths and ages are the same,then the output must be in non-decreasing alphabetical order of the names. It is guaranteed that there is no two persons share all the same of the three pieces of information. In case no one is found,output? Sample Input:12 4 Zoe_Bill 35 2333 Bob_Volk 24 5888 Anny_Cin 95 999999 Williams 30 -22 Cindy 76 76000 Alice 18 88888 Joe_Mike 32 3222 Michael 5 300000 Rosemary 40 5888 Dobby 24 5888 Billy 24 5888 Nobody 5 0 4 15 45 4 30 35 4 5 95 1 45 50 Sample Output:Case #1: Alice 18 88888 Billy 24 5888 Bob_Volk 24 5888 Dobby 24 5888 Case #2: Joe_Mike 32 3222 Zoe_Bill 35 2333 Williams 30 -22 Case #3: Anny_Cin 95 999999 Michael 5 300000 Alice 18 88888 Cindy 76 76000 Case #4: None /* Name: Copyright: Author: 流照君 Date: 2019/8/15 10:54:46 Description: */ #include <iostream> #include<string> #include <algorithm> #include <vector> #define inf 100100 using namespace std; typedef long long ll; typedef struct node{ string name; int age; int money; }person; person a[inf]; bool cmp(person x,person y) { if(x.money!=y.money) return x.money>y.money; else if(x.money==y.money&&x.age!=y.age) return x.age<y.age; else return x.name<y.name; } int main(int argc,char** argv) { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); ll n,m,num=1,l,r,num1=1,num2; cin>>n>>m; for(int i=0;i<n;i++) cin>>a[i].name>>a[i].age>>a[i].money; sort(a,a+n,cmp); while(m--) { ll flag=0; num2=1; cin>>num>>l>>r; printf("Case #%d:n",num1); for(int i=0;i<n;i++) { if(a[i].age<=r&&a[i].age>=l) { cout<<a[i].name<<" "<<a[i].age<<" "<<a[i].money<<endl; flag=1; num2++; if(num2>num) break; } } num1++; if(flag==0) cout<<"None"<<endl; } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |