A+B Problem II
时间限制:
3000 ms | 内存限制:
65535 KB
难度:
3
- 描述
-
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
A,B must be positive.
- 输入
- 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.
- 输出
- 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.
- 样例输入
-
2 1 2 112233445566778899 998877665544332211
- 样例输出
-
Case 1: 1 + 2 = 3 Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
-
这道题的解题思路很明确即用字符串,用字符串输入两个字符串,分别减去‘0’,逆序转化为两个整型数组,输出时再逆序输出即可。很简单的一道题,直接上代码!
-
#include<stdio.h> #include<algorithm> #include<math.h> #include<string.h> using namespace std; int main(){ int t,u=1; scanf("%d",&t); while(t--){ char laa[1000],lbb[1000]; //定义两个存放字符串的字符数组 int a[1000],b[1000],c[1000]; //存储相应的字符串转化为对应的十进制数 memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); scanf("%s%s",laa,lbb); int l=strlen(laa); int k=strlen(lbb); int m=0,j; j=0; for(int i=l-1;i>=0;i--){ a[m]=laa[i]-'0'; m++; } j=0; m=0; for(int i=k-1;i>=0;i--){ b[m]=lbb[i]-'0'; m++; } int p,r=0; int t=max(l,k); //取对应的l,k中的最大值 for(int i=0;i<t;i++){ //对进位的处理 p=a[i]+b[i]+r; r=p/10; c[i]=p%10; } if(r){ //这一步的作用是判断逆序后的 c[t++]++; //两个数组最后一位相加后的结果是否为1 } printf("Case %d:\n",u++); printf("%s + %s = ",laa,lbb); for(int s=t-1;s>=0;s--){ printf("%d",c[s]); } printf("\n"); } return 0; }