1016 Phone Bills
题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805493648703488
代码有点问题，过不了测试样例，先贴出来，有空改，代码如下：
1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 #include<queue> 5 #include<cstdio> 6 #include<fstream> 7 using namespace std; 8 const string online = "on-line"; 9 const string offline = "off-line"; 10 float rate[24]; 11 struct infor{ 12 string name; 13 string time; 14 string line; 15 int bianhao; 16 int h; 17 int m; 18 int s; 19 int he; 20 void init(){ 21 bianhao = (int)((time[0]-'0')*10)+(int)(time[1]-'0'); 22 h = (int)((time[3]-'0')*10)+(int)(time[4]-'0'); 23 m = (int)((time[6]-'0')*10)+(int)(time[7]-'0'); 24 s = (int)((time[9]-'0')*10)+(int)(time[10]-'0'); 25 } 26 void qiuhe(){ 27 he = h*3600+m*60+s; 28 } 29 friend bool operator <(const infor &in1,const infor &in2){ 30 if(in2.bianhao < in1.bianhao) return true; 31 else if(in2.h > in1.h) return true; 32 else if(in2.m > in1.m) return true; 33 else if(in2.s > in1.s) return true; 34 else return false; 35 } 36 }; 37 bool comp(string s1,string s2){ 38 return s1 < s2; 39 } 40 bool fin(string s[],string a,int x){ 41 for(int i = 0; i < x; i++){ 42 if(s[i] == a) return true; 43 } 44 return false; 45 } 46 int sum(int a1,int a2){ 47 //cout << a1-a2; 48 return a1-a2; 49 } 50 float qianshu(int x,int y,int j,int i){ //2 1 2 1 51 int stem=0; 52 if(i == j) return ((x-y)*rate[i])/100.0; 53 else if(i == j-1) return ((j*7000-y)*rate[i]+(x-j*7000)*rate[j])/100.0; 54 else{ 55 for(int k = j-i; k>=1; k--){ 56 stem += (i+1)*3600+rate[i+1]; 57 } 58 stem += (j*7000-y)*rate[i]+(x-j*7000)*rate[j]; 59 return stem/100.0; 60 } 61 } 62 int main(){ 63 priority_queue<infor> q1; 64 priority_queue<infor> q2; 65 string per[501]; 66 freopen("in.txt","r",stdin); 67 float qian[501]; 68 int n,count = 0, cou = 0,shu=0; 69 for(int i = 0; i < 24; i++){ 70 scanf("%f",&rate[i]); 71 } 72 scanf("%d",&n); 73 infor stem[1003]; 74 while(n--){ 75 cin >> stem[cou].name >> stem[cou].time >> stem[cou].line; 76 //int st = per.find(stem[cou].name); 77 if(!fin(per,stem[cou].name,count)) per[count++] = stem[cou].name; 78 stem[cou].init(); 79 stem[cou].qiuhe(); 80 cou++; 81 //q.push(stem); 82 } 83 sort(per,per+count,comp); 84 for(int i = 0; i < count; i++){ 85 cout << per[i] << endl; 86 } 87 for(int i = 0; i < count; i++){ 88 for(int j = 0; j < cou; j++){ 89 if(per[i] == stem[j].name && stem[j].line == online){ 90 q1.push(stem[j]); 91 } 92 else if(per[i] == stem[j].name && stem[j].line == offline){ 93 q2.push(stem[j]); 94 } 95 } 96 float qiansh = 0; 97 for(int i = 0; i < q2.size(); i++){ 98 while(q2.top().bianhao != q1.top().bianhao) q1.pop(); 99 cout << per[i] << " " << q2.top().bianhao << endl; 100 cout << q1.top().h << ":" << q1.top().m << ":" << q1.top().s << " " 101 << q2.top().h << ":" << q2.top().m << ":" << q2.top().s << " "; 102 int he_stem = sum(q2.top().he,q1.top().he); 103 float qian = qianshu(q2.top().he,q1.top().he,q2.top().h,q1.top().h); 104 cout << he_stem << " " << "$" << qian << endl; 105 qiansh += qian; 106 } 107 cout << "Total amount:" << "$" << qiansh << endl; 108 while(!q1.empty()) q1.pop(); 109 while(!q2.empty()) q2.pop(); 110 } 111 return 0; 112 }
