ACM,大数相加问题
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A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 316067 Accepted Submission(s): 61349
Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
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
解读:
四个字节的整型最多能存下10*10大小的整数,长整型也不足以存下,因此考虑用数组来存,相加之和则用整型数组存放
1、先把加数和被加数数输入到字符数组中
2、再用循环将对每位进行处理再相加
分三种情况来讨论(见代码)
切记最高位进位问题
3、最后输出
代码:
#include<iostream> #include<string.h> using namespace std; char a[1000],b[1000];//加数和被加数 int c[1000];//和 int main() { int T;//记录共有几个例子 while(scanf("%d",&T)) { int i=1;//记录当前是第几个例子 int m,n;//用来存放每个字符转化成的数字 while(T--) {//处理每个例子 int j=0;//记录和的位数 scanf("%s%s",a,b); m=strlen(a);//计算出串长 n=strlen(b); int temp=0;//用来存储进位 for(m--,n--;m>=0||n>=0;) {//逐位相加用循环, if(m>=0&&n>=0) { int x,y; x=a[m]-‘0‘;//转换成整型 y=b[n]-‘0‘; c[j]=(x+y+temp)%10; temp=(x+y+temp)/10; m--;n--;j++; } if(m>=0&&n<0) { int x; x=a[m]-‘0‘;//转换成int c[j]=(x+temp)%10; temp=(x+temp)/10; m--;j++; } if(m<0&&n>=0) { int y; y=b[n]-‘0‘; c[j]=(y+temp)%10; temp=(y+temp)/10; n--;j++; } } if(temp!=0) {//当最高位还有进位时 c[j]=temp; j++; } cout<<"Case "<<i<<":"<<endl; cout<<a<<" + "<<b<<" = "; for(;j-1>=0;j--) cout<<c[j-1]; cout<<endl; i++; if(i<T) cout<<endl; } } }
以下再给出另一种解法(毕竟思路要开阔嘛,所以就收录了接下来的这个代码)
#include<iostream> #include<string> using namespace std; char add(char temps,char tempstr,int &tempaddc) { int temp; temp = (temps - ‘0‘) + (tempstr - ‘0‘) + tempaddc; //实现进位。 tempaddc = temp / 10; // 算出进位数 return temp % 10 + ‘0‘; } int main() { string str,s; int tempaddc; char c; int n,lenstr,lens; while(cin>>n) { while(n--) { cin>>str>>s; tempaddc = 0; c = ‘0‘; if(s.length() > str.length())swap(s,str); lenstr = str.length(); lens = s.length(); lens--; lenstr--; while (lenstr >= 0) { if(lens < 0) str[lenstr] = add(c,str[lenstr],tempaddc); //字符串长度较长部分的计算 else { str[lenstr] = add(s[lens],str[lenstr],tempaddc); lens--; } lenstr--; } cout<<str<<endl; } } return 0; }
如有疑问请指出,O(∩_∩)O谢谢
至于乘法和除法之后再补充!!!
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