杭电1002——大数求和（双语博客）

Tile：HDOJ 1002 -- sum of large numbers

part one: thinking
a. Character arrays catch large numbers, attaching the numeric strings to corresponding Integer arrays(the bond arrays should be in reverse order); To get the final
two equal length large number storing in integer arrays, we fill zeros in the upper positions for the smaller one.

b. Adding each number of integer arrays coming from step a with carry, please note whether the last carry exists or not.

c. Last but not least, we must output the outputs strictly as requested.
part two: codes

See below!

1.思路

a. 字符数组接收大数，大数字符串转化为对应的int类型数组（注意字符数组和对应的整型数组应该是逆序的）；对较短的大数高位填充0，得到两个长度相同的大数整型数组。

b. 大数所在的int类型数组逐位依次相加（该进位的要进位），根据最后一步累加结果是否进位决定最终求和结果。

c. 严格按照命题格式要求输出结果。

2.代码明细

<span style="font-size:14px;">#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
int n, m;
char a[1001], b[1001];
int ai[1001], bi[1001], ci[1001];
int alen, blen, clen, maxlen, minlen;
int i, j, temp;
int num;
scanf("%d", &n);
m = n;
while(m --)
{
scanf("%s%s", &a, &b);
alen = strlen(a);
blen = strlen(b);
if(alen > blen)
{
maxlen = alen;
minlen = blen;
for(i = 0; i < alen; ++ i)
{
ai[i] = a[alen - 1 - i] - '0';
if(i < blen) {
bi[i] = b[blen - 1 - i] - '0';
} else {
bi[i] = 0;
}
}
} else {
maxlen = blen;
minlen = alen;
for(i = 0; i < blen; ++ i)
{
bi[i] = b[blen - 1 - i] - '0';
if(i < alen) {
ai[i] = a[alen - 1 - i] - '0';
} else {
ai[i] = 0;
}
}
}

j = 0;
for(i = 0; i < maxlen; ++ i) {
temp = ai[i] + bi[i] + j;
ci[i] = temp % 10;
j = temp / 10;
}
if(j > 0) {
ci[maxlen] = j;
clen = maxlen + 1;
} else {
clen = maxlen;
}

num = n - m;
printf("Case %d:\n", num);
printf("%s + %s = ", a, b);
for(i = clen - 1; i >= 0; -- i)
printf("%d", ci[i]);
if(num < n) {
printf("\n\n");
} else {
printf("\n");
}
}
return 0;
}</span>