以特定格式输出两个数的和
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
写回答
-
#include<stdio.h> #include<string.h> int main() { char a[1000],b[1000],c[1001]; int i,j=1,p=0,n,n1,n2; scanf("%d",&n); while(n--) { scanf("%s %s",a,b); printf("Case %d:\n",j); printf("%s + %s = ",a,b); n1=strlen(a)-1; n2=strlen(b)-1; for(i=0;n1>=0||n2>=0;i++,n1--,n2--) { if(n1>=0&&n2>=0){c[i]=a[n1]+b[n2]-'0'+p;} if(n1>=0&&n2<0){c[i]=a[n1]+p;} if(n1<0&&n2>=0){c[i]=b[n2]+p;} p=0; if(c[i]>'9'){c[i]=c[i]-10;p=1;} } if(p==1) printf("%d",p); while(i--) printf("%c",c[i]); j++; if(n!=1) printf("\n\n"); else printf("\n"); } }2019-07-17 19:11:28赞同 展开评论 打赏
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