HDU-2017中国大学生程序设计竞赛-网络选拔赛-1007-Palindrome Function

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ACM模版

描述

题解

枚举进制，动态规划即可。这个题是原题改的，听说是 lightoj1205 改的，原题是固定的十进制，而我们这个题进制是不固定的，需要枚举，所以在原题基础上加上枚举即可，并且注意记忆化，小心超时，初始化一次就好了。

代码

#include <iostream>#include <cstdio>using namespace std;typedef long long ll;const int MAXN = 111;const int MAXM = 50;const ll INF = 100000000000ll;int L, R, l, r, cnt;ll ans;int A[MAXN];int B[MAXN];ll dp_1[MAXM][MAXN];ll dp_2[MAXM][MAXN];void init(){    for (int i = 2; i < MAXM; i++)    {        ll mx = i;        for (int j = 1; ; j++)        {            if (j == 1)            {                dp_1[i][j] = i;                dp_2[i][j] = i;                continue;            }            int tmp = (j + 1) >> 1;            ll a = i - 1, b = i;            for (int k = 1; k < tmp; k++)            {                a *= i;                b *= i;            }            dp_1[i][j] = a;            dp_2[i][j] = b;            mx *= i;            if (mx > INF)            {                break;            }        }    }}bool check(){    for (int i = cnt - 1; i >= 0; i--)    {        if (A[i] > B[i])        {            return true;        }        if (A[i] < B[i])        {            return false;        }    }    return true;}int cal(int key){    if (cnt == 1)    {        return A[0] + 1;    }    int ret = 0;    for (int i = 1; i < cnt; i++)    {        ret += dp_1[key][i];    }    int tmp = cnt >> 1;    for (int i = cnt - 1; i >= tmp; i--)    {        int tep = cnt - (cnt - i) * 2;        int j = (i == cnt - 1) ? 1 : 0;        for (; j < A[i]; j++)        {            if (tep > 0)            {                ret += dp_2[key][tep];            }            else            {                ret++;            }        }    }    for (int i = cnt - 1; i >= tmp; i--)    {        B[i] = A[i];        B[cnt - i - 1] = A[i];    }    if (check())    {        ret++;    }    return ret;}void solve(){    ans = 0;    for (int i = l; i <= r; i++)    {        int tmp = R;        cnt = 0;        while (tmp)        {            A[cnt++] = tmp % i;            tmp /= i;        }        if (cnt == 0)        {            A[cnt++] = 0;        }        ll cnt_2 = cal(i);        tmp = L - 1;        cnt = 0;        while (tmp)        {            A[cnt++] = tmp % i;            tmp /= i;        }        if (cnt == 0)        {            A[cnt++] = 0;        }        cnt_2 -= cal(i);        ans += cnt_2 * i;        ans += (R - L + 1 - cnt_2);    }}template <class T>inline void scan_d(T &ret){    char c;    ret = 0;    while ((c = getchar()) < '0' || c > '9');    while (c >= '0' && c <= '9')    {        ret = ret * 10 + (c - '0'), c = getchar();    }}int main(){    init();    int T;    scan_d(T);    int ce = 1;    while (T--)    {        scan_d(L), scan_d(R), scan_d(l), scan_d(r);        solve();        printf("Case #%d: %lld\n", ce++, ans);    }    return 0;}

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