hdoj 1002 A + B Problem II 【大数A+B】

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 308540 Accepted Submission(s): 59647



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

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
1 2
112233445566778899 998877665544332211

[align=left]Sample Output[/align]

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
题解：大数相加，将两个字符串反转 （s1[i]-'0' s2[i]-'0'）,存于a[]  b[]，然后逐位相加，判断进位，倒序输出！！！注意格式！！！
[code]#include <cstdio>
#include <cstring>
#include <cmath>
char s1[1010],s2[1010];
int a[1010],b[1010],c[1010];
int max(int a,int b)
{
if(a>b)
return a;
else
return b;
}
int main()
{
int t,k=1;
scanf("%d",&t);
while(t--)
{

scanf("%s%s",s1,s2);
printf("Case %d:\n",k++);
printf("%s + %s = ",s1,s2);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
int la=strlen(s1);
int lb=strlen(s2);
int ans=0,cnt=0;
for(int i=la-1;i>=0;i--)
{
a[ans]=s1[i]-'0';//字符串反转
ans++;
}
for(int i=lb-1;i>=0;i--)
{
b[cnt]=s2[i]-'0';
cnt++;
}
int maxn;
maxn=max(la,lb);
for(int i=0;i<=maxn-1;i++)//从左到右逐位相加
{
a[i]=a[i]+b[i];//将各位之和还赋给数组a[]
if(a[i]>=10)//考虑各位之和大于等于10的情况，考虑进位
{
a[i]-=10;
a[i+1]++;
}
}
if(a[maxn]==0)//判断最后一位是否有进位
{
for(int i=maxn-1;i>=0;i--)
printf("%d",a[i]);
}
else
{
for(int i=maxn;i>=0;i--)
printf("%d",a[i]);
}
if(t==0)
printf("\n");
else
printf("\n\n");//每行输出间有空格
}
return 0;
}

再附上一个属于我的！用c[] 保存结果#include <cstdio>
#include <cmath>
#include <cstring>
char s1[1010],s2[1010];
int a[1010],b[1010],c[1010];
int max(int a,int b)
{
if(a>b)
return a;
else
return b;
}
int main()
{
int t,k=0;
scanf("%d",&t);
while(t--)
{

scanf("%s%s",s1,s2);
printf("Case %d:\n",k++);
printf("%s + %s = ",s1,s2);
memset(a,0,sizeof(a));//清零！！！
memset(b,0,sizeof(b));//清零 ！！！
memset(c,0,sizeof(c));//清零！！！
int la,lb;
la=strlen(s1);
lb=strlen(s2);
int ans=0,cnt=0;
for(int i=la-1;i>=0;i--)
{
a[ans]=s1[i]-'0';
ans++;
}

for(int i=lb-1;i>=0;i--)
{
b[cnt]=s2[i]-'0';
cnt++;
}
int maxn=max(la,lb);
for(int i=0;i<=maxn-1;i++)
{
if((c[i]+a[i]+b[i])>=10)
c[i+1]++;
c[i]=(c[i]+a[i]+b[i])%10;
}
if(c[maxn]==0)
{
for(int i=maxn-1;i>=0;i--)
printf("%d",c[i]);
}
else
{
for(int i=maxn;i>=0;i--)
printf("%d",c[i]);
}
if(t==0)
printf("\n");
else
printf("\n\n");
}
return 0;
} [/code]