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(杭电2053)A + B Again(转换说明符)

作者:互联网

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 39021 Accepted Submission(s): 15794

Problem Description

There must be many A + B problems in our HDOJ , now a new one is coming.
Give you two hexadecimal integers , your task is to calculate the sum of them,and print it in hexadecimal too.
Easy ? AC it !

Input

The input contains several test cases, please process to the end of the file.
Each case consists of two hexadecimal integers A and B in a line seperated by a blank.
The length of A and B is less than 15.

Output

For each test case,print the sum of A and B in hexadecimal in one line.

Sample Input

+A -A
+1A 12
1A -9
-1A -12
1A -AA

Sample Output

0
2C
11
-2C
-90

过年玩的有点嗨......emmmm....... 一开始打算用字符串储存之后一个个分析结果老是WA。。。。。。逛了一下讨论区看到了dalao们写的ac码 如下

#include <stdio.h>

int main(void)
{
    long long i,a,b,sum;

    while(scanf("%I64X %I64X",&a,&b) != EOF)
    {
        sum = a+b;
        if(sum >= 0)
        {
            printf("%I64X\n",sum);
        }
        else
        {
            sum = -sum;
            printf("-%I64X\n",sum);
        }
    }
}

发现了%x这个符号,在论坛上找出一串相关符号

(以下转自网站

转换说明符
%a(%A) 浮点数、十六进制数字和p-(P-)记数法(C99)
%c 字符
%d 有符号十进制整数
%f 浮点数(包括float和doulbe)
%e(%E) 浮点数指数输出[e-(E-)记数法]
%g(%G) 浮点数不显无意义的零"0"
%i 有符号十进制整数(与%d相同)
%u 无符号十进制整数
%o 八进制整数 e.g. 0123
%x(%X) 十六进制整数0f(0F) e.g. 0x1234
%p 指针
%s 字符串
%% "%"

2`标志
左对齐:"-" e.g. "%-20s"
右对齐:"+" e.g. "%+20s"
空格:若符号为正,则显示空格,负则显示"-" e.g. "% 6.2f"
#:对c,s,d,u类无影响;对o类,在输出时加前缀o;对x类,在输出时加前缀0x;
对e,g,f 类当结果有小数时才给出小数点。

3.格式字符串(格式)
〔标志〕〔输出最少宽度〕〔.精度〕〔长度〕类型
"%-md" :左对齐,若m比实际少时,按实际输出。
"%m.ns":输出m位,取字符串(左起)n位,左补空格,当n>m or m省略时m=n
e.g. "%7.2s" 输入CHINA
输出" CH"
"%m.nf":输出浮点数,m为宽度,n为小数点右边数位
e.g. "%3.1f" 输入3852.99
输出3853.0

标签:Again,2053,I64X,输出,浮点数,杭电,hexadecimal,sum,1A
来源: https://www.cnblogs.com/cafu-chino/p/10363306.html