标签：i++ util str iterator closed 代码 VID cst bitset
http://codeforces.com/gym/101775/problem/M
The 2018 World Cup will be hosted in Russia. 32 national teams will be divided into 8 groups. Each group consists of 4 teams. In group matches, each pair (unordered) of teams in the group will have a match. Top 2 teams with the highest score in each group will advance to eighth-finals. Winners of each eighth-final will advance to quarter-finals. Then, the winners of each quarter-final will advance to semi-finals. Eventually, the World Champion will be the winner of the World Final which is played between the two winners of the semi-finals.
Each match is labeled with a match ID sequenced from 1 to 63, with group matches followed by eighth-final matches followed by quarter-final matches followed by semi-finals matches and finally the final match.
Zhuojie is going to watch the 2018 World Cup. Since the World Champion of ACM-ICPC is very rich, he decides to spend 0.01% of his daily salary to buy tickets. However, there are only match IDs on the tickets and the prices are missing. Can you calculate how much Google pays Zhuojie every workday? Note that Zhuojie can buy multiple tickets for one match.
Input
The input starts with one line containing exactly one integer T, the number of test cases.
Each test case contains 3 lines. The first line contains 5 integers, indicating the ticket price for group match, eighth-final match, quarter-final match, semi-final match and the final match. The second line contains one integer N, the number of tickets Zhuojie buys. The third line contains N integers, each indicating the match ID on the ticket.
Output
For each test case, output one line containing "Case #x: y" where x is the test case number (starting from 1) and y is daily salary of Zhuojie.
Example
1
11 12 13 14 15
2
1 49
Case #1: 230000
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<bitset> #include<cassert> #include<cctype> #include<cmath> #include<cstdlib> #include<ctime> #include<deque> #include<iomanip> #include<list> #include<map> #include<queue> #include<set> #include<stack> #include<vector> #include <vector> #include <iterator> #include <utility> #include <sstream> #include <limits> #include <numeric> #include <functional> using namespace std; #define gc getchar() #define mem(a) memset(a,0,sizeof(a)) //#define sort(a,n,int) sort(a,a+n,less<int>()) #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); typedef long long ll; typedef unsigned long long ull; typedef long double ld; typedef pair<int,int> pii; typedef char ch; typedef double db; const double PI=acos(-1.0); const double eps=1e-6; const ll mod=1e9+7; const int inf=0x3f3f3f3f; const int maxn=1e5+10; const int maxm=100+10; bool compare(int a, int b) { return a < b;//升序 } int main() { int t = 0; int n , num; int a , b , c , d , e; int price[65] = {0}; cin >> t; for(int j = 1;j<=t;j++) { int sum = 0; scanf("%d%d%d%d%d",&a,&b,&c,&d,&e); int i; for(i=1 ;i<=48;i++) { price[i] = a; } for(i=49 ;i<=56;i++) { price[i] = b; } for(i=57 ;i<=60;i++) { price[i] = c; } for(i=61 ;i<=62;i++) { price[i] = d; } price[63] = e; scanf("%d",&n); for(i = 0;i<n;i++) { scanf("%d",&num); sum+=price[num]; } cout<<"Case #"<<j<<": "<<sum<<"0000"<<endl; } } make(ll n , ll k) { ll base = n; ll sum = 0; for(ll i = 1;i<=n;i++) { if(i >= k) { //cout<<base<<"!"<<endl;/// sum += base%1000000007; sum = sum%1000000007; } base = base*(n-i)/(1+i)%1000000007; } return sum; }
标签：i++ util str iterator closed 代码 VID cst bitset
原文地址：https://www.cnblogs.com/SutsuharaYuki/p/11950441.html
