九的余数（弃九法）
Posted meditation5201314
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Hexo博客崩了，以后不想动了，QAQ。（博客：empirefree.top）
A*B Problem
时间限制：1000 ms | 内存限制：65535 KB
难度：2
- 描述
-
设计一个程序求出A*B，然后将其结果每一位相加得到C，如果C的位数大于等于2，继续将C的各位数相加，直到结果是个一位数k。
例如：
6*8=48；
4+8=12；
1+2=3；
输出3即可。
- 输入
- 第一行输入一个数N(0<N<=1000000)，表示N组测试数据。
随后的N行每行给出两个非负整数m，n（0<=m,n<=10^12)。 - 输出
- 对于每一行数据，输出k。
- 样例输入
-
3 6 8 1234567 67 454 1232
- 样例输出
-
3 4 5
- 来源
- szhhck的水库
- 上传者
- ACM_宋志恒
- 弃九法：若一个数字的各个位数相加能被九整除，则这个数能被九整除。
- 实际应用：
- 123456789 % 9 = sum(每个位数 % 9)
- 所以：这题也是类似，每个位数相加不断%9（虽然题目没说，但是意思差不多） = 直接% 9 ，然后a * b即可.(注意a, b 一开始为0的情况)
-
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再来2道实际例题
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