Section 1.1

Your Ride Is Here

 1 /*
 2 PROG:ride
 3 LANG:C++
 4  */
 5 #include <iostream>
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <string>
 9 #include <vector>
10 #include <algorithm>
11 #include <set>
12 #include <map>
13 #include <bitset>
14 #include <cmath>
15 #include <queue>
16 #include <stack>
17 #include<fstream>
18 using namespace std;
19 string s1,s2;
20 int main()
21 {
22     ifstream fin("ride.in");
23     ofstream fout("ride.out");
24     while(fin>>s1>>s2)
25     {
26     int sum1=1,sum2=1;
27         for(int i=0;i<s1.length();i++)
28             sum1=(sum1*(s1[i]-‘A‘+1))%47;
29         for(int i=0;i<s2.length();i++)
30             sum2=(sum2*(s2[i]-‘A‘+1))%47;
31         if(sum1==sum2)
32             fout<<"GO"<<endl;
33         else
34             fout<<"STAY"<<endl;
35     }
36
37     return 0;
38 }

ride

Greedy Gift Givers

分析:模拟，开一个map维护名字和对应的收支即可，注意结束条件

 1 /*
 2 PROG:gift1
 3 LANG:C++
 4 */
 5 #include <iostream>
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <string>
 9 #include <vector>
10 #include <algorithm>
11 #include <set>
12 #include <map>
13 #include <bitset>
14 #include <cmath>
15 #include <queue>
16 #include <stack>
17 #include<fstream>
18 using namespace std;
19 const int maxn=15;
20 string x[maxn];
21 typedef pair<int,int> p;
22 map<string,p> mp;
23 int main()
24 {
25     ifstream fin("gift1.in");
26     ofstream fout("gift1.out");
27     int t;
28     fin>>t;
29     for(int i=0;i<t;i++){
30         fin>>x[i];
31         mp[x[i]].first=0; //收入
32         mp[x[i]].second=0; //支出
33     }
34     int n=t;
35     while(n--){
36         string name;
37         fin>>name;
38         int money,num;
39         fin>>money>>num;
40         if(num==0)  continue;
41         int div=money/num;
42         int mod=money%num;
43         mp[name].second+=(money-mod);
44         for(int i=0;i<num;i++){
45             string ch;
46             fin>>ch;
47             mp[ch].first+=div;
48         }
49     }
50         for(int i=0;i<t;i++){
51             fout<<x[i]<<" ";
52             fout<<mp[x[i]].first-mp[x[i]].second<<endl;
53     }
54     return 0;
55 }

gift1