题意：给定两个数的n和m，有一种操作，把 n 的各位数字加起来放到 n后面形成一个新数n，问重复 m 次所得的数能否整除 11。
析：这个题首先要知道一个规律奇数位的和减去偶数位的和能被11整除的数字一定能被11整除。当然不知道这个题也可以过，直接模拟。
还有几个其他的规律;
被3整除：每位的和能被3整除即可；
被4整除：末尾两位能被4整除即可；
被7整除：将个位数字截去，在余下的数中减去个位数字的二倍，差是7的倍数即可；（可以递归）
被8整除：末尾三位能被8整除即可；
被9整除：每位的和能被9整除即可；
被11整除：第一种方法就是用上面说的，还有一种是采用和“被7整除”一样的方法，不过要减去的是个位的一倍；
被12整除：同时被3和4整除；
被13整除：同“被7整除”，不过我们不是要减去，而是要加上个位的四倍；
被17整除：同“被7整除”，不过要减去的是个位数的五倍；
被19整除：同“被7整除”，不过要加上个位数的两倍；
代码如下：
直接模拟：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[30];
int calc(int sum, int &ans){
int cnt = 0;
while(sum){
a[cnt++] = sum % 10;
sum /= 10;
}
for(int i = cnt-1; i >= 0; --i){
ans = (ans * 10 + a[i]) % 11;
sum += a[i];
}
return sum;
}
int main(){
int kase = 0;
while(scanf("%d %d", &n, &m) == 2){
if(-1 == m && -1 == n) break;
int ans = 0;
int sum = calc(n, ans);
for(int i = 0; i < m; ++i)
sum += calc(sum, ans);
printf("Case #%d: %s\n", ++kase, ans ? "No" : "Yes");
}
return 0;
}
使用规律：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[30];
int calc(int &odd, int &even, int sum, bool &ok){
int cnt = 0;
while(sum){
a[cnt++] = sum % 10;
sum /= 10;
}
for(int i = cnt-1; i >= 0; --i, ok = !ok){
if(ok) odd += a[i];
else even += a[i];
}
return odd + even;
}
int main(){
int kase = 0;
while(scanf("%d %d", &n, &m) == 2){
if(-1 == m && -1 == n) break;
int even = 0, odd = 0;
bool ok = true;
int sum = calc(odd, even, n, ok);
for(int i = 0; i < m; ++i){
int dodd = 0, deven = 0;
sum += calc(dodd, deven, sum, ok);
odd += dodd;
even += deven;
}
int det = odd - even;
printf("Case #%d: %s\n", ++kase, det % 11 ? "No" : "Yes");
}
return 0;
}
