标签：hdoj1753

大明A+B

1.1 2.9
1.1111111111 2.3444323343
1 1.1

4
3.4555434454
2.1

import java.util.Scanner;
import java.math.BigDecimal;
public class Main{
	public static void main(String[] args){
		Scanner cin = new Scanner(System.in);
		BigDecimal a, b;
		String c;
		while(cin.hasNext()){
			a = cin.nextBigDecimal();
			b = cin.nextBigDecimal();
			c = a.add(b).stripTrailingZeros().toPlainString();
			System.out.println(c);
		}
	}
}

#include <stdio.h>
#include <string.h>
#define MAX 400 + 10
char a0[MAX], b0[MAX], s[MAX], aa[MAX], bb[MAX], ss0[MAX];  //aa保存整数，a保存小数

int main(){
    char *dot;
    while( scanf( "%s%s", a0, b0 ) == 2 ){
        char *a = a0, *b = b0, *ss = ss0;
        aa[0] = bb[0] = ‘0‘;  //留着进位
        dot = strchr( a, ‘.‘ );
        if( dot == NULL ){ strcpy( aa + 1, a ); a[0] = ‘\0‘; } //如果a是整数，将a置空
        else{ dot[0] = ‘\0‘; strcpy( aa + 1, a ); a = dot + 1; }
        dot = strchr( b, ‘.‘ );
        if( dot == NULL ){ strcpy( bb + 1, b ); b[0] = ‘\0‘; }
        else{ dot[0] = ‘\0‘; strcpy( bb + 1, b ); b = dot + 1; }
        
        //接下来开始计算大整数相加aa+bb
        int aai = strlen( aa ), bbi = strlen( bb );  //保存整数位数
        int ssi = aai > bbi ? aai : bbi;
        ss[ssi] = ‘\0‘;
        int i = aai < bbi ? aai : bbi;  //保存较小位数
        int sign = 0, t; //保存进位
        while( i-- ){
            t = aa[--aai] - ‘0‘ + ( bb[--bbi] - ‘0‘ );
            t += sign; sign = 0;
            if( t > 9 ) { ++sign; t -= 10; }
            ss[--ssi] = t + ‘0‘;
        }
        while( aai ){ 
            t = aa[--aai] - ‘0‘;  //例如1+99999的情况
            t += sign; sign = 0;
            if( t > 9 ) { ++sign; t -= 10; }
            ss[--ssi] = t + ‘0‘; 
        }
        while( bbi ){ 
            t = bb[--bbi] - ‘0‘; 
            t += sign; sign = 0; 
            if( t > 9 ) { ++sign; t -= 10; }
            ss[--ssi] = t + ‘0‘; 
        }
        //大整数加法aa+bb完成，2014-1-25 00:59:04
        
        //小数部分的加法，注意位要对齐，sign已被化成0，2014-1-25 09:49:02
        int ai = strlen( a ), bi = strlen( b );  //保存小数位数
        int si = ai > bi ? ai : bi;
        s[si] = ‘\0‘;
        int j = ai < bi ? ai : bi;  //保存较小位数
        while( j-- ){
            t = a[j] - ‘0‘ + ( b[j] - ‘0‘ );
            t += sign; sign = 0;
            if( t > 9 ){ ++sign; t -= 10; }
            s[j] = t + ‘0‘;
        }
        j = ai < bi ? ai : bi;
        if( ai < bi ) strcpy( s + j, b + j ); 
        else if( ai > bi ) strcpy( s + j, a + j );        
        
        //将小数位可能产生的进位整合到整数和ss中
        ssi = strlen( ss );
        if( sign ){ //如果小数相加产生了个位进位1
            while( ssi ){
                t = ss[--ssi] - ‘0‘ + sign;
                sign = 0;
                if( t > 9 ){ t -= 10; ++sign; }
                ss[ssi] = t + ‘0‘;
            } //整数部分整合完毕            
        }
        
        //处理小数和的末尾0
        si = strlen( s );
        while( si ){
            if( s[--si] == ‘0‘ ){ s[si] = ‘\0‘; continue; }
            break;
        }
        
        //处理整数和可能存在的前导0
        int ii = 0;
        for( ; ss[ii] != ‘\0‘; ++ii ){
            if( ss[ii] == ‘0‘ ) continue;
            break;
        }
        ss = ss + ii;
        if( !strlen(ss) ){  //防止ss被减成空字符串
            ss[0] = ‘0‘;
            ss[1] = ‘\0‘;
        }
        //输出结果
        printf( "%s", ss );
        si = strlen( s );
        if( si ){
            putchar( ‘.‘ );
            printf( "%s", s );
        }
        putchar( ‘\n‘ );
    }
    return 0;
}

HDU 1753 大明A+B

标签：hdoj1753

原文地址：http://blog.csdn.net/chang_mu/article/details/24595109