Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem D: HST

On July 1st, Ontario's Provincial Sales Tax (PST) was merged with the Federal Goods and Services Tax (GST), creating the Harmonized Sales Tax (HST). This changed the rate of tax Ontarians pay on various items.

Dalton would like to calculate how this change will affect his personal monthly budget.

Input Specification

The first line of input contains a single integer, the number of test cases to follow. Each test case begins with a line containing two integers N, M, the number of categories of items and the number of purchases Dalton makes each month, respectively. Each of these integers is between 1 and 100000, inclusive. N lines follow, each describing a category of item. Each of these lines contains a category name, which is a string of at most 30 uppercase letters, followed by three percentages, giving the PST, GST, and HST rate on that category. Each percentage is a number with up to two digits after the decimal point, and each percentage is followed by the % symbol. Each percentage is at least 0% and at most 100%. These lines are followed by M more lines, each describing one of Dalton's purchases. Each of these lines contains a category name and a price in dollars and cents and prefixed by the $ symbol. Each such line indicates the amount of pre-tax money that Dalton spends on an item in the specified category. The amount of each tax to be paid on an item is rounded to the nearest cent; if the tax amount is exactly half a cent more than a whole number of cents, it is rounded up to the nearest greater whole number of cents.

Sample Input

1
1 1
gas 0% 5% 13%
gas $100.00

Output Specification

For each test case, output the difference, in dollars and cents, between the total HST payable and the total sum of PST and GST payable on Dalton's monthly purchases. If the HST is more than the PST+GST, output a positive amount. If the HST is less than the PST+GST, output a negative amount.

Output for Sample Input

8.00

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題目描述：

由於將兩種稅收(PST, GST) 合併成一個新的 HST 稅收制度，給定這三種稅收格式，
問新的制度與舊的制度的稅收差值。

題目解法：

每個稅率百分比最多 2 位，為了防止小數點誤差，考慮將所有數字儲存成整數。
而每筆花費也是最多 2 位，對於 0 位 1 位要特別處理。

而計算單筆消費時採用四捨五入計算稅收。

#include <stdio.h>
#include <string.h>
#include <map>
#include <iostream>
using namespace std;
struct E {
    int PST, GST, HST;
    E(int a, int b, int c):
        PST(a), GST(b), HST(c) {}
    E():
        PST(0), GST(0), HST(0) {}
};
int readPercent() {
    //most two digit after '.', try a.bb => a*100+bb
    char s[100], s2[100];
    scanf("%s", s);
    int a, b;
    if(sscanf(s, "%d.%s", &a, &s2) == 2) {
        sscanf(s2, "%d", &b);
        if(strlen(s2) == 2)
            a = a*100 + b*10;
        else
            a = a*100 + b;
    } else {
        sscanf(s, "%d", &a);
        a = a*100;
    }
    return a;
}
long long readDollar() {
    //most two digit after '.', try a.bb => a*100+bb
    char s[100], s2[100];
    scanf("%s", s);
    long long a, b;
    if(sscanf(s, "$%lld.%s", &a, &s2) == 2) {
        sscanf(s2, "%lld", &b);
        if(strlen(s2) == 1)
            a = a*100 + b*10;
        else
            a = a*100 + b;
    } else {
        sscanf(s, "$%lld", &a);
        a = a*100;
    }
    return a;
}
long long calcTax(long long n, long long p) {
    long long tax = n*p;//n/100 * p/100 round up
    return tax/10000 + (tax%10000 >= 5000);
}
int main() {
    int testcase;
    int n, m;
    int i, j, k;
    char name[105], s[105];
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d", &n, &m);
        map<string, E> R;
        int pst, gst, hst;
        for(i = 0; i < n; i++) {
            scanf("%s", name);
            pst = readPercent();
            gst = readPercent();
            hst = readPercent();
            R[name] = E(pst, gst, hst);
        }
        long long ret = 0;
        for(i = 0; i < m; i++) {
            scanf("%s", name);
            long long val = readDollar();
            ret -= calcTax(val, R[name].PST);
            ret -= calcTax(val, R[name].GST);
            ret += calcTax(val, R[name].HST);
        }
        if(ret < 0) printf("-"), ret = -ret;
        printf("%lld.%02lld\n", ret/100, ret%100);
    }
    return 0;
}