hdu 5690 2016"百度之星" - 初赛（Astar Round2A） All X 快速二次幂 || 寻找周期

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5690

题意：m个数字全为x mod k ?= c;其中m <= 10¹⁰,0 < c,k <= 10,000;

法1：xxx = (10^m-1)/9*x;但是n太大，需要同时mod。去除分母将式子变为：10^m*x%(9k) - x%(9k) =? 9c ;其中 10^m 快速二次幂即可；

时间复杂度为O(logn)

 

法2： 由于m个x数的产生对于mod具有可拆分性，所以直接求解周期即可；

#include<bits/stdc++.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1|1
typedef pair<int,int> PII;
#define A first
#define B second
#define MK make_pair
typedef long long ll;
typedef unsigned int uint;
template<typename T>
void read1(T &m)
{
    T x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>9) out(a/10);
    putchar(a%10+'0');
}
const int maxn = 1e4+7;
int mp[maxn], rk[maxn];
int main()
{
    //freopen("data.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T, kase = 1;
    scanf("%d",&T);
    while(T--){
        MS1(mp);
        ll x, m, k, c, circle = -1, aux = 0, start;
        read2(x,m); read2(k,c);
        rep1(i,1,m){
            aux = (aux*10+x)%k;
            rk[i] = aux;
            if(mp[rk[i]] != -1){
                circle = i - mp[rk[i]];
                start = mp[rk[i]];
                break;
            }
            mp[rk[i]] = i;
        }
        //printf("\n%lld %lld\n\n",start,circle);
        printf("Case #%d:\n",kase++);
        if(circle == -1)　　　　
            puts(aux == c?"Yes":"No");
        else{
            int ans = rk[(m - start)%circle + start];
            puts(ans == c?"Yes":"No");
        }
    }
    return 0;
}