大数+找规律 ACdream1210 Chinese Girls' Amusement

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传送门：点击打开链接

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;
}

大数+找规律 ACdream1210 Chinese Girls' Amusement

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原文地址：http://blog.csdn.net/qwb492859377/article/details/47981571