1669 - A+B without carry
Xiao Ming always tends to ignore the carry when he does decimal addition with paper and pencil.For example,15+17,Xiao Ming will answer 22,because he ignores the carry from the single digits.5+7=12,and the digit 1 is the carry.
题目输入
The input will consist of a series of pairs of integers a and b(both less than 1000000000),separated by a space, one pair of integers per line.
题目输出
For each pair of input integers a and b you should output the correct answer of the sum of a and b,a space character and Xiao Ming's answer of the sum of a and b in one line,and with one line of output for each line in input.If Xiao Ming's answer begins with zero,don't output unnecessary zero.
输入/输出样例
题目输入
15 16 1 999 31 71
题目输出
31 21 1000 990 102 2
C语言解答
#include <stdio.h> #include <math.h> void compute(int a,int b) { int m,n,i,j,p=1; i=a+b; j=i; while((a/10!=0)||(b/10!=0)) { m=a%10;n=b%10; if(m+n>=10) j=j-(int)pow(10,p); a=a/10;b=b/10; p++; } if((a%10)+(b%10)>=10) j=j-(int)pow(10,p); printf("%d ",i); printf("%d\n",j); } int main() { int a,b; while(scanf("%d %d",&a,&b)!=EOF) compute(a,b); return 0; }
C++解答
#include <stdio.h> int a,b; void run() { int c,k; c=a+b; printf("%d ",c);//这个程序是由正常的和来反求不进位的和（或者直接按位相加也是可以的 ）。 k=1; while(k<=a||k<=b) { if(a/k%10+b/k%10>9) c-=k*10;//在某一位上的进位被忽略，相当于总和减小了这一位权值的十倍，个位进位被忽略，总和就少了10。 k*=10; } printf("%d\n",c); } int main() { while(scanf("%d%d",&a,&b)!=EOF) run(); return 0; }