九的余数（弃九法）

　　Hexo博客崩了，以后不想动了，QAQ。（博客：empirefree.top）

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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 70                                                                         .                                                                             **/
 71                                                                         
 72 #include 
 73 #include 
 74 #include 
 75 #include 
 76 #include <string.h>
 77 #include 
 78 #include 
 79 #include 
 80 using namespace std;
 81 
 82 #define FFI(a, b) for(int i = a; i  83 #define FFJ(a, b) for(int j = a; j  84 #define FFK(a, b) for(int k = a; k  85 #define RR(a, b) for(int i = a; i > b; i++)
 86 #define ME(a, b) memset(a, b, sizeof(a))
 87 #define SC(x) scanf("%d", &x)
 88 #define SCC(a, b) scanf("%d%d", &a, &b)
 89 #define PR(x) printf("%d\n", x)
 90 #define PRR(a, b) printf("%d%d\n", a, b);
 91 #define INF 0x3f3f3f3f
 92 #define MAX 202
 93 #define MOD 1000000007
 94 #define E 2.71828182845
 95 #define M 8
 96 #define N 6
 97 typedef long long LL;
 98 const double PI = acos(-1.0);
 99 typedef pair<int, int> Author;
100 vectorstring, int> > VP; 
101 
102 /*******************************************************************************************************************************************************/
103 
104 int main(void){
105 //    #ifdef Empirefree
106 //        freopen("in.txt", "r", stdin);
107 //        freopen("out.txt", "w", stdout);
108 //    #endif
109 //    
110 //    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
111     int n;LL a, b, s;
112     
113     SC(n);
114     //弃九法 
115     while(n--){
116         
117         scanf("%lld%lld", &a, &b); 
118         if(!a || !b){
119             printf("0\n");
120             continue;
121         }
122         s = (a % 9) * (b % 9) % 9;
123         
124         printf("%lld\n", s == 0 ? 9 : s);
125     }
126     
127     return EXIT_SUCCESS;
128 }        

Empirefree.top

再来2道实际例题

http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=205

http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=105