【t014】拯数

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【题目链接】:http://noi.qz5z.com/viewtask.asp?id=t014

【题意】

【题解】

这个锁的序列,如果把末尾的0去掉;然后再倒过来;那么就是这个序列对应的格雷码了;然后根据格雷码和二进制的对应关系;能够处理出对应的二进制;比如格雷码存在a数组,二进制存在b数组;则b[1] = a[1]rep1(i,2,n)   b[i] = b[i-1]^a[i];然后把这个二进制转成十进制就是答案了要写高精度哦

【完整代码】

#include <cstdio>#include <algorithm>#include <iostream>#include <cmath>#include <cstring>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define rei(x) scanf("%d",&x)#define rel(x) scanf("%lld",&x)typedef pair<int, int> pii;typedef pair<LL, LL> pll;const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };const double pi = acos(-1.0);const int N = 1100;struct abc{    int len, a[N];    abc()    {        memset(a, 0, sizeof a);    }};int n,a[N],b[N];abc two[N], ans;abc plu(abc a, abc b){    int len = max(a.len, b.len);    abc c;    int x = 0;    rep1(i, 1, len)    {        c.a[i] = a.a[i] + b.a[i] + x;        x = c.a[i] / 10;        c.a[i] %= 10;    }    while (x)    {        c.a[++len] = x;        c.a[len] %= 10;        x /= 10;    }    c.len = len;    return c;}int main(){    //freopen("F:\\rush.txt", "r", stdin);    rei(n);    rep1(i, 1, n)        cin >> a[i];    while (n - 1 >= 1 && a[n] == 0) n--;    reverse(a + 1, a + 1 + n);    b[1] = a[1];    rep1(i, 2, n)        b[i] = b[i - 1] ^ a[i];    rep1(i, 1, n)        a[i] = b[i];    two[0].len = 1, two[0].a[1] = 1;    rep1(i, 1, n)    {        two[i].len = two[i - 1].len;        rep1(j, 1, two[i].len)            two[i].a[j] = two[i - 1].a[j];        int &len = two[i].len,x = 0;        rep1(j, 1, len)        {            two[i].a[j] = two[i].a[j] * 2 + x;            x = two[i].a[j] / 10;            two[i].a[j] %= 10;        }        while (x > 0)        {            two[i].a[++len] = x;            two[i].a[len] %= 10;            x = x / 10;        }    }    ans.a[1] = 0, ans.len = 1;    rep2(i, n, 1)    if (a[i])        ans = plu(ans, two[n - i]);    rep2(i, ans.len, 1)        printf("%d", ans.a[i]);    puts("");    return 0;}

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