HPU 1002 A + B Problem II【大數】
阿新 • • 發佈：2017-07-03
lar 保存 memory positive test size mod ack auth
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
Sample Output
Author Ignatius.L
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 261413 Accepted Submission(s): 50581
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
Author Ignatius.L
思路：
將數字以字符的形式存儲到字符數組中，由於在存儲的時候是高位在以0為下標的下標變量中存儲的。所以要將其進行翻轉，存儲到整形數組中（也就是高位存儲到大下標變量中。由於在進位的時候能在原來的基礎上進行i++。來存儲最高位的數據）。然後將兩個大數按位相加。假設比十大，進行進位操作！代碼：
#include <stdio.h>
#include <string.h>
#define N 10005
char a[N],b[N];
int c[N],d[N];
int main()
{
int n,i,j,k,len1,len2;
scanf("%d",&n);
k=n;
while(n--)
{
memset(c,0,sizeof(c));//每次都得清零，所以得放到while循環裏面。
memset(d,0,sizeof(d));
getchar();
scanf("%s%s",a,b);//空格也是scanf的切割符！
len1=strlen(a);
len2=strlen(b);
for(i=len1-1,j=0;i>=0;i--)//由於須要逆序保存，所以應該設變量j從0開始！
c[j++]=a[i]-'0';
for(i=len2-1,j=0;i>=0;i--)
d[j++]=b[i]-'0';
for(i=0;i<1001;i++)
{
c[i]+=d[i];
if(c[i]>=10)
{
c[i]-=10;
c[i+1]++;
}
}
printf("Case %d:\n%s + %s = ",k-n,a,b);
for(i=1000;i>=0&&c[i]==0;i--);
if(i>=0)
for(;i>=0;i--)
{
printf("%d",c[i]);
}
else
printf("0");
printf("\n");
if(n!=0)
printf("\n");
}
return 0;
}
HPU 1002 A + B Problem II【大數】