HDU 6796 X Number(数位、指数型母函数)
X Number
Time Limit: 3000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 631 Accepted Submission(s): 272
Problem Description
Teitoku loves many different kinds of numbers, and today Little W wants him to classify some positive integers into different categories.
There are 11 categories, numbered from 0 to 10. For each positive integer x, if there exists only one type of digit d (0≤d≤9) that occurs in x with the highest frequency, then we say x should be classified into category d, or otherwise, in case such digit doesn't exist, we say x should be classified into category 10.
For example,
● 111223 should be classified into category 1 since digit 1 occurs three times, and digits 2 and 3 occur less than three times respectively, and
● 3345544 should be classified into category 4 since digit 4 occurs three times, and digits 3 and 5 occur less than three times respectively, and
● 112233 should be classified into category 10 since digits 1, 2 and 3 occur twice respectively.
Little W doesn't care about category 10 and he just wants Teitoku to tell him the number of integers ranged from l to r that should be classified to another category d. However, Teitoku can hardly solve this problem, so he asks you for help.
There are 11 categories, numbered from 0 to 10. For each positive integer x, if there exists only one type of digit d (0≤d≤9) that occurs in x with the highest frequency, then we say x should be classified into category d, or otherwise, in case such digit doesn't exist, we say x should be classified into category 10.
For example,
● 111223 should be classified into category 1 since digit 1 occurs three times, and digits 2 and 3 occur less than three times respectively, and
● 3345544 should be classified into category 4 since digit 4 occurs three times, and digits 3 and 5 occur less than three times respectively, and
● 112233 should be classified into category 10 since digits 1, 2 and 3 occur twice respectively.
Little W doesn't care about category 10 and he just wants Teitoku to tell him the number of integers ranged from l to r that should be classified to another category d. However, Teitoku can hardly solve this problem, so he asks you for help.
Input
There are several test cases.
The first line contains an integer T (1≤T≤1000), denoting the number of test cases. Then follow all the test cases.
For each test case, the only line contains three integers l, r and d (1≤l≤r≤1018,0≤d≤9), representing a problem.
The first line contains an integer T (1≤T≤1000), denoting the number of test cases. Then follow all the test cases.
For each test case, the only line contains three integers l, r and d (1≤l≤r≤1018,0≤d≤9), representing a problem.
Output
For each test case, output in one line the number of integers ranged from l to r that should be classified to category d.
Sample Input
3 1 10 1 1 11 1 1 100 0
Sample Output
1 2 1
Hint
For the sample cases, 1 and 11 are in category 1, 100 is in category 0 and 10 is in category 10.
Source
2020 Multi-University Training Contest 3
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题意:
一个数字当数位中出现次数最多的数字是z时，且唯一最多，则这个数字属于z类。
思路:
类似于数位dp，计算到某个数位，计算到当前数位有限制和非限制情况。当非限制时
后面可以任意选数字，我们可以使用指数型母函数来计算方案数。
对于这题当计算到某一位，没限制且没前导零，可以直接计算。
比如计算1000内属于1的数字，由数字正着第一位开始。
第一位选了0，不满足能直接计算的条件，继续向下展开，第二位选到0了，这个时候仍不满足直接计算条件，
继续向下展开...第二位选1的时候，这个时候满足计算条件，因为后两位可以任意选，这个时候计算已经含有1个数字1，且
后面两位任意选的方案数，通过指数型母函数可以算的此时的方案数为19种，第二位选2~9时，每一种方案都是1，共8种。
当第二位为0向下展开的时候，到了第三位，这个时候如果第三位仍然为0会继续向下展开，而第三位选1的时候
即剩余1位，要求1为众数，这个时候计算的方案数为1，第三位选2~9的方案数为0。第三位选0继续向下到第四位的
时候如果选1是一种方案，选2~9没有方案。再回到第一层，第一位选1的时候到了第二位虽然没有前导零，但是仍
有lim限制，所以继续展开，一直到最后一位的时候都没有能直接计算，到了第0位，check一下z是不是最多的数字。
总和就是19+8+1+1=29种。
这样直接计算不记忆化的话，大概是5s，记忆化的时候不需要考虑前面选的哪些数字，只需要这些的数字的个数
比如z = 3， 前面选了2个1、1个2跟前面选了1个2，2个1的计算结果一样，并不会影响到3的摆法。
存入vector，用一个map套vector就可以了，时间只需要265ms。
#include
#include
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e3 + 500;
const int M = 4e5 + 10;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
#ifdef LOCAL
#define TIME cout << "RuningTime: " << clock() << "ms\n", 0
#else
#define TIME 0
#endif
map, ll>mp;
int x[20];
int a[20];
long double g[20], f[20];
int z;
int cnt;
long double fac[20];
void init()
{
fac[0] = 1;
for (int i = 1; i <= 18; ++i)
{
fac[i] = fac[i - 1] * i;
}
}
ll check()
{
for (int i = 0; i <= 9; ++i)
{
if (i != z)
{
if (a[i] >= a[z])
return 0;
}
}
return 1;
}
ll calc(int pos)
{
int mx = 0;
for (int i = 0; i <= 9; ++i)
{
if (i != z)
mx = max(mx, a[i]);
}
ll tot = 0;
int MI = max(mx + 1, a[z]) - a[z]; //最少摆
int MX = pos; //最多摆
for (int m = MI; m <= MX; ++m) //枚举后面pos个位置摆了多少z
{
memset(f, 0, sizeof f);
f[0] = 1;
for (int i = 0; i <= 9; ++i) //指数型母函数求解
{
memset(g, 0, sizeof g);
int begin = i == z ? m : 0; //强制z选了m个，其余的可以不选
int end = i == z ? m : min(pos - (m - a[z]), m + a[z] - a[i] - 1); //选其他的数字不能数目超过z
for (int j = 0; j <= pos; ++j)
{
for (int k = begin; k <= end && j + k <= pos; ++k)
g[j + k] += f[j] / fac[k];
}
memcpy(f, g, sizeof f);
}
tot += f[pos] * fac[pos] + 0.5; //精损
}
return tot;
}
ll dfs(int pos, int lim, int zero)
{
ll sum = 0;
if (pos == 0)
{
return check();
}
if (!lim && zero) //没限制且没前导零，可以直接计算
{
vectorv;
for (int i = 0; i <= 9; ++i) //将信息存储用于记忆化
{
if (i != z && a[i])
v.push_back(a[i]);
}
sort(v.begin(), v.end());
v.push_back(a[z]);
v.push_back(pos);
if (mp.find(v) != mp.end())
return mp[v];
return mp[v] = calc(pos);
}
int up = lim ? x[pos] : 9;
for (int i = 0; i <= up; ++i) //需要继续展开的情况
{
if (zero || i != 0) //前导零不加0的个数
a[i]++;
sum += dfs(pos - 1, lim && i == up, zero || i);
if (zero || i != 0)
a[i]--;
}
return sum;
}
ll calc(ll num)
{
cnt = 0;
while (num)
{
x[++cnt] = num % 10;
num /= 10;
}
return dfs(cnt, 1, 0);
}
int main()
{
#ifdef LOCAL
freopen("E:\\input.txt", "r", stdin);
freopen("E:\\output.txt", "w", stdout);
#endif
init();
int t;
cin >> t;
while (t--)
{
ll l, r;
scanf("%lld%lld%d", &l, &r, &z);
printf("%lld\n", calc(r) - calc(l - 1));
}
return TIME;
}