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

描述

题解

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;
}