1168: Bill (pointer topic)

1168: bill （ Pointer theme ）

Title Description
At the end of every month , Xiao Ming will sort out and count the expenditure bill of this month . Nowadays, computers have become popular on university campuses , So Xiao Ming wants the computer to help with this . Smart, just make a program for Xiao Ming to complete this thing .
Input
Multi instance testing . First, enter an integer ncase, Indicates the number of test instances . The input for each test case is as follows ：
The first line is an integer n (n<100). And then there was n Line billing information , Each line consists of the name of the thing name And the corresponding cost c form , Length not exceeding 200. There will be one or more spaces in the middle , And there are no spaces at the beginning and end of each line . 0.0 < c < 1000.0.
Output
Each test instance corresponds to one line of output , Total cost of output , Keep one digit after the decimal point .
The sample input Copy
2
1
Buy books 62.28
3
Apple 2.3
Buy clothes for girl friend 260.5
Go to cinema 30
Sample output Copy
62.3
292.8
source / classification
Method 1

#include<stdio.h>
#include<string.h>
#define N 200
/*  Accurately find the cost of each item   Find the subscript of the first space in reverse order , Again +1, Is where spending begins   for example ,56.3 It's a string , First calculate 563, Because there is only one decimal place ,563 Except in 10 that will do  */

int main(){
    
    char *str,*c;
    int ncase,n,f=0;
    double sum=0,money=0,count=1;
    str=(char *)malloc(sizeof(char)*N);
    scanf("%d",&ncase);

    while(ncase--){
    
        scanf("%d ",&n);
        sum=0;// Set the total amount to 0
        while(n--){
    
            gets(str);
            count=1;// Record the number of digits after the decimal point 
            money=0;
            f=0;
            // The pointer c Point to price 
            for(int i=strlen(str)-1;i>=0;i--){
    
                if(str[i]==' ') {
    
                    c=&str[i+1];
                    break;
                }
            }
            // take char conversion double
            for(int i=0;c[i]!='\0';i++){
    
                if(c[i]!='.') money=money*10+(c[i]-'0');
                if(c[i]=='.') f=1;// Marked with a decimal point 
            }
			// If there is a decimal point, divide it by the corresponding number of digits 
            if(f==1){
    
                for(int i=strlen(c)-1;i>=0;i--){
    
                    if(c[i]=='.') break;
                    count*=10;
                }
            }
            sum+=money/count;
        }
        printf("%.1lf\n",sum);
    }
    return 0;
}

Method 2

#include<stdio.h>
#include<string.h>
#define N 200
/*  Skillfully using function  strrchr() and sscanf() */

int main(){
    
    char *str,*c;
    int ncase,n;
    double sum=0,money=0;
    str=(char *)malloc(sizeof(char)*N);
    scanf("%d",&ncase);

    while(ncase--){
    
        scanf("%d ",&n);
        sum=0;// Set the total amount to 0
        while(n--){
    
            gets(str);
            c=strrchr(str,' ');// Go back to the last ' ' stay str The subscript , No return NULL
            sscanf(c,"%lf",&money);// If successful, return the number of parameters , Failure returns -1.
            sum+=money;
        }
        printf("%.1lf\n",sum);
    }
    return 0;
}