[codeforces 516]A. Drazil and Factorial
[codeforces 516]A. Drazil and Factorial
试题描述
Drazil is playing a math game with Varda.
Let's define 
for positive integer x as a product of factorials of its digits. For example, 
.
First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions:
1. x doesn't contain neither digit 0 nor digit 1.
2. = 
.
Help friends find such number.
输入
The first line contains an integer n (1 ≤ n ≤ 15) — the number of digits in a.
The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes.
输出
输入示例
4 1234
输出示例
33222
数据规模及约定
见“输入”
题解
这个题相当有意思，我看 n 那么小，那一定是 dp 了，然而写完了才发现 F(a) 会爆 long long，所以只好另辟蹊径。后来发现一个性质：只要把 a 的每一位数都尽量的分出最多数出来，然后再拼到一起就好了，这个不难证明，就是个贪心，若将两个数的阶乘合并成一个数的阶乘，则答案会减少 1，一定不优。
现在的任务是把 2, 3, 4, ... , 9 这些 1 位数尽量多地分解，我发现刚刚的 dp 没有白写：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
if(Head == Tail) {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
return *Head++;
}
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
#define maxn 6000010
#define maxk 20
int n, sum, fact[maxk], F[maxn], f[maxn][11];
char num[maxk];
bool vis[maxn];
bool Less(int* a, int* b) {
for(int i = 9; i >= 2; i--) if(a[i] != b[i])
return a[i] < b[i];
return 0;
}
int main() {
n = read();
scanf("%s", num + 1);
fact[0] = 1;
for(int i = 1; i <= 9; i++) fact[i] = fact[i-1] * i;
sum = 1;
for(int i = 1; i <= n; i++) sum *= fact[num[i]-'0'];
vis[1] = 1;
for(int i = 1; i <= sum; i++) if(vis[i]) {
// printf("%d ", i);
for(int k = 2; k <= 9 && fact[k] * i <= sum; k++) {
int t = fact[k] * i;
// printf("[%d]", t);
vis[t] = 1;
if(F[t] < F[i] + 1) {
F[t] = F[i] + 1;
memcpy(f[t], f[i], sizeof(f[i]));
f[t][k]++;
}
if(F[t] > F[i] + 1) continue;
f[i][k]++;
if(Less(f[t], f[i])) memcpy(f[t], f[i], sizeof(f[i]));
f[i][k]--;
}
}
for(int i = 9; i >= 2; i--)
for(int j = 1; j <= f[sum][i]; j++)
putchar(i + '0');
putchar('\n');
return 0;
}
就依次输入，结果记一下，在主程序中打个表：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
if(Head == Tail) {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
return *Head++;
}
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
#define maxn 20
int n, cnt[maxn];
char num[maxn];
int main() {
n = read();
scanf("%s", num + 1);
for(int i = 1; i <= n; i++) {
if(num[i] == '9') {
cnt[7]++; cnt[3]++; cnt[3]++; cnt[2]++;
}
if(num[i] == '8') {
cnt[7]++; cnt[2]++; cnt[2]++; cnt[2]++;
}
if(num[i] == '7') {
cnt[7]++;
}
if(num[i] == '6') {
cnt[5]++; cnt[3]++;
}
if(num[i] == '5') {
cnt[5]++;
}
if(num[i] == '4') {
cnt[3]++; cnt[2]++; cnt[2]++;
}
if(num[i] == '3') {
cnt[3]++;
}
if(num[i] == '2') {
cnt[2]++;
}
}
for(int i = 7; i >= 2; i--)
for(int j = 1; j <= cnt[i]; j++) putchar(i + '0');
putchar('\n');
return 0;
}
A 啦！
