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GDUFE ACM-1002
A+B(Big Number Version)
Time Limit: 2000/1000ms (Java/Others)
Problem Description:
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 400.
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:
3 1 2 112233445566778899 998877665544332211 33333333333333333333333333 100000000000000000000
Sample Output:
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110 Case 3: 33333333333333333333333333 + 100000000000000000000 = 33333433333333333333333333
这道题难度中等
1 核心代码: 2 scanf("%s%s",s1,s2); 3 len1=strlen(s1); len2=strlen(s2); 4 for(i=len1-1,j=0;i>=0;i--) //将s1字符串数组转换为数字数组,逆序 5 num1[j++]=s1[i]-‘0‘; 6 for(i=len2-1,j=0;i>=0;i--) 7 num2[j++]=s2[i]-‘0‘; 8 for(i=0;i<N;i++) 9 { 10 num1[i]+=num2[i]; 11 if(num1[i]>9) //进位 12 { 13 num1[i]-=10; //加法最大是9+9=18 14 num1[i+1]++; //所以加1 15 } 16 } 17 printf("Case %d:\n%s + %s = ",X,s1,s2); 18 X++; 19 for(i=N-1;(i>=0)&&(num1[i]==0);i--); //空语句去掉多余的0 20 for(;i>=0;i--) 21 printf(“%d”,num1[i]); //逆序输出结果 22
另附一种AC代码:
1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 char a[1000],b[1000],c[1001]; 6 int i,j=1,p=0,n,n1,n2; 7 scanf("%d",&n); 8 while(n) 9 { 10 scanf("%s %s",a,b); 11 printf("Case %d:\n",j); 12 printf("%s + %s = ",a,b); 13 n1=strlen(a)-1; 14 n2=strlen(b)-1; 15 for(i=0;n1>=0||n2>=0;i++,n1--,n2--) 16 { 17 if(n1>=0&&n2>=0){c[i]=a[n1]+b[n2]-‘0‘+p;} 18 if(n1>=0&&n2<0){c[i]=a[n1]+p;} 19 if(n1<0&&n2>=0){c[i]=b[n2]+p;} 20 p=0; 21 if(c[i]>‘9‘){c[i]=c[i]-10;p=1;} 22 } 23 if(p==1){printf("%d",p);} 24 while(i--) 25 {printf("%c",c[i]);} 26 j++; 27 if(n!=1){printf("\n\n");} 28 else {printf("\n");} 29 n--; 30 } 31 }
GDUFE ACM-1002
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