HDU-1002 A + B Problem II
2012-02-17 21:09
323 查看
[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
这个题 开个存数的数组 a[1001],然后把两串相加的结果存入a中 再对a中大于10的元素进行操作
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
这个题 开个存数的数组 a[1001],然后把两串相加的结果存入a中 再对a中大于10的元素进行操作
#include<stdio.h> #include<string.h> int main() { char str1[1001],str2[1001],t[1001]; int n,f = 0,a[1001],i,j,flag,k1,k2,k,x,flag1; scanf("%d", &n); while(n--) { f++;//这个f是为了后面输出case 1。。。用的 memset(a,0,sizeof(a));//将a中全初始化为0 k = 0;//k是a的下标 最初为0 flag = 0; flag1 = 0; scanf("%s %s", str1,str2); k1 = strlen(str1); k2 = strlen(str2); if(k1<k2)//让str1始终为大的 { flag = 1; strcpy(t,str1); strcpy(str1,str2); strcpy(str2,t); x = k1;//把长度也交换下 k1 = k2; k2 = x; } for(i = k1-1,j = k2-1 ; j>= 0 ;i--,j--)//开始从后往前加 { a[k] = a[k]+str1[i]-'0'+str2[j]-'0';//因为是字符 -‘0’这里要注意是a[k]+因为a[k]有可能本来就有值 if(a[k]>=10)//把和存入k之后 判断一下k是否是大于10的 { a[k+1] += a[k]/10;//把进上去的一位 存入a[k+1]中 a[k] = a[k]%10;//把各位留给a[k],这里两个语句不可相反 k++; } else k++; } for(i = k1-k2-1 ; i >= 0 ; i--)//如果str1还没完的话 按照上面依次存入a中 { a[k] = a[k]+str1[i]-'0'; if(a[k]>=10) { a[k+1] += a[k]/10; a[k] = a[k]%10; k++; } else k++; } while(a[k] == 0)//把前面的0全筛掉 k--; printf("Case %d:\n",f); if(flag == 1) printf("%s + %s = ",str2,str1); else printf("%s + %s = ",str1,str2); for(i = k; i >= 0 ; i--) { printf("%d",a[i]); flag1 = 1;//如果有输出 就标记下 } if(flag1 == 0)//如果没输出 就说明全都是0 就是 0 0 相加了 直接输出 0 printf("0"); printf("\n"); if(n!=0) printf("\n"); } return 0; }
相关文章推荐
- hdu-1002 A + B Problem II
- HDU 1002 A + B Problem II (BigNums)
- hdu 1002 A + B Problem II(高精度加法)
- ACM:HDU-1002 A + B Problem II
- hdu 1002 A+B problem II
- HDU 1002 A + B Problem II (大数加法)
- hdu1002 A + B Problem II
- HDU 1002 -- A+B problemII (Java)
- [HDU1002] A + B Problem II
- HDU 1002 A + B Problem II (Java 之大整数)
- hdu 1002 A + B Problem II
- HDU 1002 A + B Problem II 大数相加
- hdu1002 A + B Problem II(大数相加)
- hdu 1002 A + B Problem II(大正整数相加)
- HDU 1002 A+B problemII
- hdu 1002 A + B Problem II(大数相加)
- HDU 1002 A + B Problem II(高精度)
- hdu 1002 A + B Problem II
- hdu 1002:A + B Problem II(大数问题)
- 大数A + B Problem II-hdu-1002