- 题目描述:
-
Children are taught to add multi-digit numbers from right-to-left one digit at a time. Many find the "carry" operation - in which a 1 is carried from one digit position to be added to the next - to be a significant challenge. Your job is to count the number of carry operations for each of a set of addition problems so that educators may assess their difficulty.
- 输入:
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Each line of input contains two unsigned integers less than 10 digits. The last line of input contains 0 0.
- 输出:
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For each line of input except the last you should compute and print the number of carry operations that would result from adding the two numbers, in the format shown below.
- 样例输入:
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123 456 555 555 123 594 0 0
- 样例输出:
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NO carry operation. 3 carry operations. 1 carry operation.
如果是NO,最后是没有s的,如果是1,也是没有s的。其他才是复数形式。万万注意。
Code:#include <iostream> using namespace std; void Translate(int arr[],unsigned int num,unsigned int &cnt){ cnt=0; while(num!=0){ int x=num%10; arr[cnt++]=x; num=num/10; } } int main() { int num_a[20],num_b[20]; unsigned int a,len_a,b,len_b; unsigned int index_a,index_b; while(cin>>a>>b){ if(a==0&&b==0) break; Translate(num_a,a,len_a); Translate(num_b,b,len_b); index_a=index_b=0; int answer=0,c=0; while(index_a<len_a&&index_b<len_b){ c=(num_a[index_a]+num_b[index_b]+c)/10; if(c>0) ++answer; ++index_a; ++index_b; } if(index_a<len_a){ while(index_a<len_a){ c=(num_a[index_a]+c)/10; if(c>0) ++answer; ++index_a; } } if(index_b<len_b){ while(index_b<len_b){ c=(num_b[index_b]+c)/10; if(c>0) ++answer; ++index_b; } } if(answer==0) cout<<"NO carry operation."<<endl; else if(answer==1) cout<<answer<<" carry operation."<<endl; else cout<<answer<<" carry operations."<<endl; } return 0; } /************************************************************** Problem: 1143 User: lcyvino Language: C++ Result: Accepted Time:0 ms Memory:1520 kb ****************************************************************/