ZOJ 2812 Quicksum

Quicksum

题目传送门~

Time Limit: 2000 msMemory Limit: 65536 KB

A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change, so checksums are often used for detecting transmission errors, validating document contents, and in many other situations where it is necessary to detect undesirable changes in data.

For this problem, you will implement a checksum algorithm called Quicksum. A Quicksum packet allows only uppercase letters and spaces. It always begins and ends with an uppercase letter. Otherwise, spaces and letters can occur in any combination, including consecutive spaces.

A Quicksum is the sum of the products of each character's position in the packet times the character's value. A space has a value of zero, while letters have a value equal to their position in the alphabet. So, A=1, B=2, etc., through Z=26. Here are example Quicksum calculations for the packets "ACM" and "MID CENTRAL":

        ACM: 1*1  + 2*3 + 3*13 = 46

MID CENTRAL: 1*13 + 2*9 + 3*4 + 4*0 + 5*3 + 6*5 + 7*14 + 8*20 + 9*18 + 10*1 + 11*12 = 650

Input: The input consists of one or more packets followed by a line containing only # that signals the end of the input. Each packet is on a line by itself, does not begin or end with a space, and contains from 1 to 255 characters.

Output: For each packet, output its Quicksum on a separate line in the output.

Example Input:	Example Output:
ACM MID CENTRAL REGIONAL PROGRAMMING CONTEST ACN A C M ABC BBC #	46 650 4690 49 75 14 15

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题后两句话：

①注意输入输出的组织（五何问题之 何时开始，何时结束，何时输入，何时输出，何时变量初始化！）

②这题最关键的是如何读取带有空格的字符串，由此引出两种方法gets()和cin,getline()

③这题我学到了：for循环的结束条件要活学活用！

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方法一：c++——cin,getline()

#include<stdio.h>
int main()
{
	char str[256];
	int sum,i;
	while(gets(str)&&str[0]!='#'){
		sum=0;     		
		for(i=0;str[i]!='\0';i++)
			if(str[i]!=' ') sum+=(i+1)*(str[i]-'@');	
		printf("%d\n",sum);
	}
	return 0;
}

方法二：c——gets()

#include<stdio.h>
int main()
{
	char str[256];
	int sum,i;
	while(gets(str)&&str[0]!='#'){
		sum=0;     		
		for(i=0;str[i]!='\0';i++)
			if(str[i]!=' ') sum+=(i+1)*(str[i]-'@');	
		printf("%d\n",sum);
	}
	return 0;
}