PAT甲级 1016 Phone Bills (25 分)

最新推荐文章于 2024-09-17 20:21:58 发布

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本文介绍了一种在线离线计费系统的实现方案，通过分析设备的在线和离线时间，精确计算出每个设备的使用费用。系统采用C++语言进行编程，详细解析了时间戳转换、费用计算及数据结构的运用。

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很久没写模拟题了，，写的很蠢

马一下，待整理。。。ZZZ

#include<bits/stdc++.h>
#define FI first
#define SE second

using namespace std;
typedef long long ll;
typedef pair<string, string> P;
const int maxn = 100 + 7;

/*
分析：

CYLL 01:01:06:01 on-line
CYLL 01:01:08:03 off-line
2 * 60 + 2 = 122  *  X  =  24.40
6 59    1180
7 60    1200
8 3     60

CYLL 01:28:15:41 on-line
CYLL 01:28:16:05 off-line



CYJJ 01:01:05:59 on-line
CYJJ 01:01:07:00 off-line



aaa 01:01:01:03 on-line

aaa 01:02:00:01 on-line
aaa 01:04:23:59 off-line

aaa 01:05:02:24 on-line


*/
int a[maxn];
int n, m;
map<string, vector<P> > mp;
struct node{
    string na, tm, tp;
}x[1007];
bool cmp(node a, node b) {
    if(a.na == b.na)
        return a.tm < b.tm;
    else return a.na < b.na;
}

double work(string s1, string s2) {
    cout << s1 << " " << s2;
    int cnt = 0;
    double res = 0;
    int d1 = ( (s1[0] - '0')*10 + (s1[1] - '0') );
    int h1 = ( (s1[3] - '0')*10 + (s1[4] - '0') );
    int m1 = ( (s1[6] - '0')*10 + (s1[7] - '0') );
    int d2 = ( (s2[0] - '0')*10 + (s2[1] - '0') );
    int h2 = ( (s2[3] - '0')*10 + (s2[4] - '0') );
    int m2 = ( (s2[6] - '0')*10 + (s2[7] - '0') );

    if((d1 == d2) && (h1 == h2)) {
        cnt += (m2 - m1);
        res += (cnt * a[h1]);
        cout << " " << cnt << " $";
        printf("%.2f\n", res/100.0);
        return res;
    }
    else if(d1 == d2) {
        if(m1 > 0) {
            cnt += (60 - m1);
            res += ((60 - m1) * a[h1]);
            h1++;
            if(h1 == 24) {
                h1 = 0;
                d1++;
            }
        }
        cnt += m2;
        res += (m2 * a[h2]);
        h2--;
        if(h2 == -1) {
            h2 = 23;
            d2--;
        }
        for(int i = h1; i <= h2; ++i) {
            cnt += 60;
            res += (60 * a[i]);
        }
        cout << " " << cnt << " $";
        printf("%.2f\n", res/100.0);
        return res;
    }
    else {
        if(m1 > 0) {
            cnt += (60 - m1);
            res += ((60 - m1) * a[h1]);
            h1++;
            if(h1 == 24) {
                h1 = 0;
                d1++;
            }
        }
        cnt += m2;
        res += (m2 * a[h2]);
        h2--;
        if(h2 == -1) {
            h2 = 23;
            d2--;
        }

        if(d1 == d2) {
            for(int i = h1; i <= h2; ++i) {
                cnt += 60;
                res += (60 * a[i]);
            }
        }
        else if(d1 < d2) {
            d1++;
            for(int i = h1; i <= 23; ++i) {
                cnt += 60;
                res += (60 * a[i]);
            }
            d2--;
            for(int i = 0; i <= h2; ++i) {
                cnt += 60;
                res += (60 * a[i]);
            }
            for(int i = d1; i <= d2; ++i) {
                cnt += (24 * 60);
                res += (60 * m);
            }
        }
        cout << " " << cnt << " $";
        printf("%.2f\n", res/100.0);
        return res;
    }
}
void solve(string na) {
    sort(mp[na].begin(), mp[na].end());
    double sum = 0;
    for(int i = 0; i < mp[na].size()-1; ++i) {
        if(mp[na][i].SE == "on-line" && mp[na][i+1].SE == "off-line") {
            sum += work(mp[na][i].FI.substr(3, 8), mp[na][i+1].FI.substr(3, 8));
        }
    }
    printf("Total amount: $%.2f\n", sum/100.0);
}
int main() {
    m = 0;
    for(int i = 0; i <= 23; ++i) {
        scanf("%d", &a[i]);
        m += a[i];
    }
    scanf("%d", &n);
    string na, tm, tp;
    for(int i = 1; i <= n; ++i) {
        cin >> x[i].na >> x[i].tm >> x[i].tp;
        //mp[na].push_back(make_pair(tm, tp));
    }
    sort(x+1, x+1+n, cmp);
    for(int i = 1; i < n; ++i) {
        if((x[i].na == x[i+1].na) && (x[i].tp == "on-line" && x[i+1].tp == "off-line")) {
            mp[x[i].na].push_back(make_pair(x[i].tm, x[i].tp));
            mp[x[i+1].na].push_back(make_pair(x[i+1].tm, x[i+1].tp));
        }
    }
    for(auto i : mp) {
        cout << i.FI ;
        cout << " " << (i.SE)[0].FI[0];
        cout << i.SE[0].FI[1] << endl;
        solve(i.FI);
    }

    return 0;
}