大数+找规律 ACdream1210 Chinese Girls' Amusement

来源：互联网发布：java 乐观锁用法编辑：程序博客网时间：2024/05/06 11:23

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题意：对于n个点围成的圈。从一个点出发，顺时针数K个位置，一直进行这个操作直到回到最初的那个点时，恰好把所有的点都访问了一遍，问最大的K(K<=n/2)

思路：很容易就想到了一种方法，找到K<=n/2，且gcd(K,n)=1，有人是用java从n/2向1去枚举的，感觉好暴力，所以当时不敢这样写

后来发现其实是有规律的，从n=3一直算下去，会得到一个这样的序列1 1 2 1 3 3 4 3 5 5 6 5 7 7 8 7 9 9 10 9.....

很明显以4个为一组，一下子就能找到规律。。

然后按照这个规律直接算出答案就行了

#include<map>#include<set>#include<cmath>#include<stack>#include<queue>#include<cstdio>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define FIN freopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w+",stdout)using namespace std;typedef long long LL;typedef pair<int, int> PII;const int MX = 2500;const int MAXN = 9999;const int DLEN = 4;class Big {public:    int a[MX], len;    Big(const int b = 0) {        int c, d = b;        len = 0;        memset(a, 0, sizeof(a));        while(d > MAXN) {            c = d - (d / (MAXN + 1)) * (MAXN + 1);            d = d / (MAXN + 1);            a[len++] = c;        }        a[len++] = d;    }    Big(const char *s) {        int t, k, index, L, i;        memset(a, 0, sizeof(a));        L = strlen(s);        len = L / DLEN;        if(L % DLEN) len++;        index = 0;        for(i = L - 1; i >= 0; i -= DLEN) {            t = 0;            k = i - DLEN + 1;            if(k < 0) k = 0;            for(int j = k; j <= i; j++) {                t = t * 10 + s[j] - '0';            }            a[index++] = t;        }    }    Big operator/(const int &b)const {        Big ret;        int i, down = 0;        for(int i = len - 1; i >= 0; i--) {            ret.a[i] = (a[i] + down * (MAXN + 1)) / b;            down = a[i] + down * (MAXN + 1) - ret.a[i] * b;        }        ret.len = len;        while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;        return ret;    }    bool operator>(const Big &T)const {        int ln;        if(len > T.len) return true;        else if(len == T.len) {            ln = len - 1;            while(a[ln] == T.a[ln] && ln >= 0) ln--;            if(ln >= 0 && a[ln] > T.a[ln]) return true;            else return false;        } else return false;    }    Big operator+(const Big &T)const {        Big t(*this);        int i, big;        big = T.len > len ? T.len : len;        for(i = 0; i < big; i++) {            t.a[i] += T.a[i];            if(t.a[i] > MAXN) {                t.a[i + 1]++;                t.a[i] -= MAXN + 1;            }        }        if(t.a[big] != 0) t.len = big + 1;        else t.len = big;        return t;    }    Big operator-(const Big &T)const {        int i, j, big;        bool flag;        Big t1, t2;        if(*this > T) {            t1 = *this;            t2 = T;            flag = 0;        } else {            t1 = T;            t2 = *this;            flag = 1;        }        big = t1.len;        for(i = 0; i < big; i++) {            if(t1.a[i] < t2.a[i]) {                j = i + 1;                while(t1.a[j] == 0) j++;                t1.a[j--]--;                while(j > i) t1.a[j--] += MAXN;                t1.a[i] += MAXN + 1 - t2.a[i];            } else t1.a[i] -= t2.a[i];        }        t1.len = big;        while(t1.a[t1.len - 1] == 0 && t1.len > 1) {            t1.len--;            big--;        }        if(flag) t1.a[big - 1] = 0 - t1.a[big - 1];        return t1;    }    int operator%(const int &b)const {        int i, d = 0;        for(int i = len - 1; i >= 0; i--) {            d = ((d * (MAXN + 1)) % b + a[i]) % b;        }        return d;    }    Big operator*(const Big &T) const {        Big ret;        int i, j, up, temp, temp1;        for(i = 0; i < len; i++) {            up = 0;            for(j = 0; j < T.len; j++) {                temp = a[i] * T.a[j] + ret.a[i + j] + up;                if(temp > MAXN) {                    temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);                    up = temp / (MAXN + 1);                    ret.a[i + j] = temp1;                } else {                    up = 0;                    ret.a[i + j] = temp;                }            }            if(up != 0) {                ret.a[i + j] = up;            }        }        ret.len = i + j;        while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;        return ret;    }    void print() {        printf("%d", a[len - 1]);        for(int i = len - 2; i >= 0; i--) printf("%04d", a[i]);    }};int main() {    char word[MX];    while(~scanf("%s", word)) {        Big n(word);        Big a = (n - 3) / 4 + 1;        Big b = a * 2 - 1;        Big c = n - a * 4 + 2;        if(c.a[0] == 3) (b + 1).print();        else b.print();        printf("\n");    }    return 0;}

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