hdu 1002 A + B Problem II 菜鸟求帮助

Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

Sample Input
2
1 2
112233445566778899 998877665544332211

Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

#include<stdio.h>
#include<string.h>
int main()
{
	int n,i=0,j;
	scanf("%d",&n);
	while(i<n)
	{
		char a[1000],b[1000];
	    int sum[1002]={0},la,lb,q,save;
		i++;
	    scanf("%s%s",&a,&b);
		la=save=strlen(a);
		lb=strlen(b);
		q=la>lb?la:lb;
		for(j=1;j<=la;j++)
			sum[j]=a[j-1]-"0";
		if(la>=lb)   
		{
			for(j=q;lb>0;j--,lb--)
				sum[j]+=(b[lb-1]-"0");
		}
		else 
		{
			for(j=q;la>0;j--,lb--,la--)
				sum[j]+=(b[lb-1]-"0"+a[la-1]-"0");
			for(j;lb>0;lb--,j--)
				sum[j]=(b[lb-1]-"0");
		}
		for(j=q+1;j>0;j--)
		{
			if(sum[j]>=10)
			{
				sum[j-1]++;
				sum[j]-=10;
			}
		}
		printf("Case %d:\n%s + %s = ",i,a,b);
		if(sum[0]!=0)printf("%d",sum[0]);
		for(j=1;j<q+1;j++)
		{
			printf("%d",sum[j]);
		}
		if(i!=n)printf("\n\n");
		else printf("\n");
	}
	return 0;
}

解决方案

参考:

/*
 * hdu-1002
 * mike-w
 * 2012-5-21
 */
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define ONLINE_JUDGE
#ifndef ONLINE_JUDGE
#include<assert.h>
#endif
#define MAX_NUM_LEN 1234
#define BASE 10000
#define WIDTH 4
#define min(a,b) ((a)>(b)?(b):(a))
int conv(char *buf, int *s)
{
	int len=strlen(buf);
	int weight=1,pos=MAX_NUM_LEN-1;
	int i;
	memset(s,0,sizeof(int)*MAX_NUM_LEN);
	for(i=1;i<=len;i++)
	{
		s[pos]+=(buf[len-i]-"0")*weight;
		weight*=10;
		if(weight==BASE)
			weight=1, pos--;
	}
	s[0]=pos;
	return 0;
}
int add(int *a1, int *a2, int *s)
{
	memset(s,0,sizeof(int)*MAX_NUM_LEN);
	int end=min(a1[0], a2[0]);
	int carry=0,i;
	for(i=MAX_NUM_LEN-1;i>=end;i--)
	{
		s[i]=a1[i]+a2[i]+carry;
		carry=s[i]/BASE;
		s[i]%=BASE;
	}
	if(carry)
		s[i]=carry,s[0]=i;
	else
		s[0]=i+1;
	return 0;
}
int disp(int *s)
{
	int i=s[0];
	while(i<MAX_NUM_LEN && s[i]==0)
		i++;
	if(i==MAX_NUM_LEN)
		putchar("0");
	else
		printf("%d",s[i]);
	for(i++;i<MAX_NUM_LEN;i++)
		printf("%0*d",WIDTH,s[i]);
	return 0;
}
int read_num(int *s)
{
	char buf[MAX_NUM_LEN];
	memset(s,0,sizeof(int)*MAX_NUM_LEN);
	scanf("%s",buf);
	conv(buf,s);
	return 0;
}
int main(void)
{
#ifndef ONLINE_JUDGE
	assert(freopen("in","r",stdin));
#endif
	int ncase,ccase;
	int n1[MAX_NUM_LEN];
	int n2[MAX_NUM_LEN];
	int s[MAX_NUM_LEN];
	scanf("%d",&ncase);
	for(ccase=1;ccase<=ncase;ccase++)
	{
		if(ccase>1)
			putchar("\n");
		printf("Case %d:\n",ccase);
		read_num(n1);
		read_num(n2);
		add(n1, n2, s);
		disp(n1)
		printf(" + ");
		disp(n2);
		printf(" = ");
		disp(s);
		printf("\n");
	}
	return 0;
}
/*
 * extra words:
 * 本人的存储大数的方法：
 * char* -> int*
 * 数字在数组中与尾部对齐——不倒序存储！
 * 数组的首位标示了数字可能开始的位置，有可能比实际开始位置
 * 提前。
 * 输出的过程中注意过滤掉数组前面的0
 */

仅供参考：

#include <stdio.h>
#include <string.h>
#define MAXLEN 1000
char a1[MAXLEN];
char a2[MAXLEN];
static int v1[MAXLEN];
static int v2[MAXLEN];
static int v3[MAXLEN];
int i,j,n,L,z;
void main(void) {
    scanf("%d",&n);
    for (j=0;j<n;j++) {
        scanf("%s%s",a1,a2);
        L=strlen(a1);
        for (i=0;i<L;i++) v1[i]=a1[L-1-i]-"0";
        L=strlen(a2);
        for (i=0;i<L;i++) v2[i]=a2[L-1-i]-"0";
        for (i=0;i<MAXLEN;i++) v3[i]=v1[i]+v2[i];
        for (i=0;i<MAXLEN;i++) {
            if (v3[i]>=10) {
                v3[i+1]+=v3[i]/10;
                v3[i]=v3[i]%10;
            }
        }
        printf("Case %d:\n", j+1);
        printf("%s + %s = ", a1, a2);
        z=0;
        for (i=MAXLEN-1;i>=0;i--) {
            if (z==0) {
                if (v3[i]!=0) {
                    printf("%d",v3[i]);
                    z=1;
                }
            } else {
                printf("%d",v3[i]);
            }
        }
        if (z==0) printf("0");
        printf("\n");
    }
}
//Sample Input
//3
//0 0
//1 2
//112233445566778899 998877665544332211
//
//Sample Output
//Case 1:
//0 + 0 = 0
//Case 2:
//1 + 2 = 3
//Case 3:
//112233445566778899 + 998877665544332211 = 1111111111111111110