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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 879 Accepted Submission(s): 421
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<string>
#include<iostream>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
using namespace std;
typedef long long LL;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
#pragma comment(linker, "/STACK:102400000,102400000")
const int maxn = 1e5 + 300;
const LL INF = 0x3f3f3f3f;
typedef long long LL;
typedef unsigned long long ULL;
LL vis[maxn], a[maxn];
int main(){
LL x, m, k, c;
int T, cas = 0;
scanf("%d",&T);
while(T--){
scanf("%lld%lld%lld%lld",&x,&m,&k,&c);
memset(vis,0,sizeof(vis));
int nn = 0, le = 0, st = 1;
LL cc;
LL n = 0;
for(int i = 1; ; i++){
n = n*10;
n = n + x;
n = n%k;
if(vis[n]){
cc = n;
break;
}else{
vis[n] = 1;
nn++;
a[nn] = n;
}
}
for(int i = 1; i <= nn; i++){
if(a[i] == cc){
le = nn+1 - i;
st = i;
break;
}
}
int flag = 0;
if(m < st){
if(a[m] == c){
flag = 1;
}
}else{
m -= st;
m %= le;
if(a[st+m] == c){
flag = 1;
}
}
printf("Case #%d:\n",++cas);
if(flag) puts("Yes");
else puts("No");
}
return 0;
}
/*
55
3 5 99 69
3 4 4 2
3 8 4 2
2 8 3 2
*/
解题思路：数学方法。
转自http://m.blog.csdn.net/article/details?id=51471639
这个数要mod k ,那这个数应该怎么表示呢?
就这样转化了,然后10^m可以通过快速幂解决,但是很明显,除以9操作怎么办,除法取模,余数是会改变的,逆元?但是9和k不一定互质,且k也不一定是质数,所以扩展欧几里得和费马小定理都不能用了,束手无策
好吧,这里提供一种小方法
就这样经过几步转化,我求d不需要进行除法取模了,那我上面的问题不就解决了?对的。
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<string>
#include<iostream>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
using namespace std;
typedef long long LL;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
#pragma comment(linker, "/STACK:102400000,102400000")
const int maxn = 1e5 + 300;
const LL INF = 0x3f3f3f3f;
typedef long long LL;
typedef unsigned long long ULL;
LL quick(LL x,LL n,LL p){
if(n == 0) return 1;
LL ret = 1;
while(n){
if(n&1) ret = ret*x % p;
n = n >> 1;
x = x*x % p;
}
return ret;
}
int main(){
LL x, m, k, c;
int T, cas = 0;
scanf("%d",&T);
while(T--){
scanf("%lld%lld%lld%lld",&x,&m,&k,&c);
k*=9;
LL ans = quick(10,m,k);
ans = (ans - 1 + k) % k;
ans /= 9;
k /= 9;
ans = ans * x % k;
bool flag = 0;
if(ans == c)
flag = 1;
printf("Case #%d:\n",++cas);
if(flag) puts("Yes");
else puts("No");
}
return 0;
}
HDU 5690——All X——————【快速幂 | 循环节】
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原文地址：http://www.cnblogs.com/chengsheng/p/5521041.html
