nyist A+B Problem IV

A+B Problem IV

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 3

描述

acmj最近发现在使用计算器计算高精度的大数加法时很不方便，于是他想着能不能写个程序把这个问题给解决了。

输入

包含多组测试数据
每组数据包含两个正数A,B（可能为小数且位数不大于400）

输出

每组输出数据占一行，输出A+B的结果，结果需要是最简的形式。

样例输入

1.9 0.1
0.1 0.9
1.23 2.1
3 4.0

样例输出

2
1
3.33
7

字符串处理
考虑0这个特殊的数

 #include <stdio.h>
#include <string.h>

#define N 905

char ch_1[N],ch_2[N],ch1_1[N],ch1_2[N];
char ch3[N],ch4[N];
char ch5[N],ch6[N];/*整数加  和小数后的加*/
void f()
{
    ch_1[0] = '\0';
    ch_2[0] = '\0';
    ch1_1[0] = '\0';
    ch1_2[0] = '\0';
    ch5[0] = '\0';
    ch6[0] = '\0';
}
void judge_char(char ch[],char ch1[],char ch2[])
{
    ch1[0] = '\0';
    ch2[0] = '\0';

    int i = 0;
    for(i = 0; i < strlen(ch); i++)
        if(ch[i] == '.')
            break;

    int j = 0,k  = 0;

    for(j = 0; j < i; j++)
            ch1[j] = ch[j];
            ch1[j++] = '\0';

    int t = 0;
    for(k = i+1; k < strlen(ch); k++)
        {
            ch2[t++] = ch[k];
        }
        ch2[t] = '\0';
}

void add_int(char ch[],char ch1[],char ch2[],int kk)/*ch为长度短着  整数相加*/
{
    ch2[0] = '\0';

    int i = 0,j = 0;
    int k = strlen(ch1)-1;
    int t = kk;
    int t1 = 0;
    int x = 0;
    for(i = strlen(ch)-1; i >= 0; i--)
    {
        x = (int) ch[i] + (int)ch1[k] - 2*'0' + t;
        t = x /10;
        x = x % 10;
        ch2[t1++] = x + '0';
        k--;
    }

    for(j = k; j >= 0; j--)
    {
    
        x = (int)ch1[j] +t - '0';
        t = x /10;
        x = x % 10;
        ch2[t1++] = x + '0';

    }

    if(t)
    {
        
        ch2[t1++] =(char)(t + '0');
        t = 0;
    }
    ch2[t1++] = '\0';
}

int add_float(char ch[],char ch1[],char ch2[])/*小数相加 ch为长度短着*/
{
    ch2[0] = '\0';

    int i = 0;
    int j = 0;
    int k = 0;
    int t = 0;
    int t1 = 0;
    int x = 0;

    for(i = strlen(ch1) - 1; i >= strlen(ch); i--)
    {
        x = (int)(ch1[i] + t - '0');
		t = x/10;
		x %= 10;
        ch2[t1++] =(char)(x + '0');

    }

    for(j = strlen(ch)-1; j >=0 ; j--)
    {
        x = (int)(ch[j] + ch1[j] + t - 2*'0');
		t = x /10;
		x = x % 10;
        ch2[t1++] = (char)(x + '0');
    }

    ch2[t1++] = '\0';
    return t;
}

int main()
{
    while(scanf("%s%s",ch3,ch4) !=EOF)
    {
        f();
        int kk = 0;

        judge_char(ch3,ch_1,ch_2);
        judge_char(ch4,ch1_1,ch1_2);
        
        if(ch_2[0] == '\0')
            {
                ch_2[0] = '0';
                ch_2[1] = '\0';
            }
        if(ch1_2[0] == '\0')
        {
            ch1_2[0] = '0';
            ch1_2[1] = '\0';
        }

        if(strlen(ch_2) > strlen(ch1_2))
            kk = add_float(ch1_2,ch_2,ch6);
        else
            kk = add_float(ch_2,ch1_2,ch6);

        if(strlen(ch_1) > strlen(ch1_1) )
            add_int(ch1_1,ch_1,ch5,kk);
        else
            add_int(ch_1,ch1_1,ch5,kk);

		 int i = 0;
		 int xx = 0;
		 int yy = 0;
		 for(i = 0;i <= strlen(ch6)-1; i++)
            if(ch6[i] == '0')
                xx++;
        if(xx == strlen(ch6) )
            {
                yy = strlen(ch5)-1;

                for(i = yy; i >= 0; i-- )
                    if(ch5[i] != '0') break;

                while(yy >= 0)
                {
                    printf("%c",ch5[yy]);
                    yy--;
                }
                printf("\n");
            }
        else
        {
                yy = strlen(ch5)-1;
                while(yy >= 0)
                {
                    printf("%c",ch5[yy]);
                    yy--;
                }
                printf(".");
                yy = strlen(ch6)-1;
                i = 0;
                while(i <= yy)
                    {
                        if(ch6[i] != '0')
                        break;
                        i++;
                    }

                while(yy >= i)
                {
                    printf("%c",ch6[yy]);
                    yy--;
                }
                printf("\n");
        }


    }
    return 0;

}