【HDOJ】3419 The Three Groups

记忆化搜索。

 1 /* 3419 */
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5
 6 #define MAXN 10
 7
 8 int ans;
 9 int n[3];
10 int x[3];
11 bool visit[MAXN];
12 int cnt[MAXN][MAXN][MAXN];
13
14 void dfs(int d, int i, int v) {
15     if (i == 3) {
16         if (x[0]*x[1] == x[2])
17             ++ans;
18         return ;
19     }
20     if (d < n[i]) {
21         for (int j=1; j<MAXN; ++j) {
22             if (!visit[j]) {
23                 visit[j] = true;
24                 dfs(d+1, i, v*10+j);
25                 visit[j] = false;
26             }
27         }
28     } else {
29         x[i] = v;
30         dfs(0, i+1, 0);
31     }
32 }
33
34 int main() {
35     int a, b, c;
36     int i, j, k;
37
38     #ifndef ONLINE_JUDGE
39         freopen("data.in", "r", stdin);
40         freopen("data.out", "w", stdout);
41     #endif
42
43     memset(cnt, -1, sizeof(cnt));
44     while (scanf("%d %d %d", &a, &b, &c)!=EOF && (a||b||c)) {
45         if (a+b < c) {
46             puts("0");
47         } else if (a>c || b>c){
48             puts("0");
49         } else if (a+b-c>1) {
50             puts("0");
51         } else if (cnt[a][b][c] >= 0) {
52             printf("%d\n", cnt[a][b][c]);
53         } else {
54             memset(visit, false, sizeof(visit));
55             n[0] = a;
56             n[1] = b;
57             n[2] = c;
58             ans = 0;
59             dfs(0, 0, 0);
60             cnt[a][b][c] = ans;
61             printf("%d\n", ans);
62         }
63     }
64
65     return 0;
66 }
时间: 2024-11-04 20:54:03

【HDOJ】3419 The Three Groups的相关文章

【HDOJ】4956 Poor Hanamichi

基本数学题一道,看错位数,当成大数减做了,而且还把方向看反了.所求为最接近l的值. 1 #include <cstdio> 2 3 int f(__int64 x) { 4 int i, sum; 5 6 i = sum = 0; 7 while (x) { 8 if (i & 1) 9 sum -= x%10; 10 else 11 sum += x%10; 12 ++i; 13 x/=10; 14 } 15 return sum; 16 } 17 18 int main() { 1

【HDOJ】1099 Lottery

题意超难懂,实则一道概率论的题目.求P(n).P(n) = n*(1+1/2+1/3+1/4+...+1/n).结果如果可以除尽则表示为整数,否则表示为假分数. 1 #include <cstdio> 2 #include <cstring> 3 4 #define MAXN 25 5 6 __int64 buf[MAXN]; 7 8 __int64 gcd(__int64 a, __int64 b) { 9 if (b == 0) return a; 10 else return

【HDOJ】2844 Coins

完全背包. 1 #include <stdio.h> 2 #include <string.h> 3 4 int a[105], c[105]; 5 int n, m; 6 int dp[100005]; 7 8 int mymax(int a, int b) { 9 return a>b ? a:b; 10 } 11 12 void CompletePack(int c) { 13 int i; 14 15 for (i=c; i<=m; ++i) 16 dp[i]

【HDOJ】3509 Buge&#39;s Fibonacci Number Problem

快速矩阵幂,系数矩阵由多个二项分布组成.第1列是(0,(a+b)^k)第2列是(0,(a+b)^(k-1),0)第3列是(0,(a+b)^(k-2),0,0)以此类推. 1 /* 3509 */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #incl

【HDOJ】1818 It&#39;s not a Bug, It&#39;s a Feature!

状态压缩+优先级bfs. 1 /* 1818 */ 2 #include <iostream> 3 #include <queue> 4 #include <cstdio> 5 #include <cstring> 6 #include <cstdlib> 7 #include <algorithm> 8 using namespace std; 9 10 #define MAXM 105 11 12 typedef struct {

【HDOJ】2424 Gary&#39;s Calculator

大数乘法加法,直接java A了. 1 import java.util.Scanner; 2 import java.math.BigInteger; 3 4 public class Main { 5 public static void main(String[] args) { 6 Scanner cin = new Scanner(System.in); 7 int n; 8 int i, j, k, tmp; 9 int top; 10 boolean flag; 11 int t

【HDOJ】2425 Hiking Trip

优先级队列+BFS. 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 using namespace std; 6 7 #define MAXN 25 8 9 typedef struct node_st { 10 int x, y, t; 11 node_st() {} 12 node_st(int xx, int yy, int tt)

【HDOJ】1686 Oulipo

kmp算法. 1 #include <cstdio> 2 #include <cstring> 3 4 char src[10005], des[1000005]; 5 int next[10005], total; 6 7 void kmp(char des[], char src[]){ 8 int ld = strlen(des); 9 int ls = strlen(src); 10 int i, j; 11 12 total = i = j = 0; 13 while (

【HDOJ】2795 Billboard

线段树.注意h范围(小于等于n). 1 #include <stdio.h> 2 #include <string.h> 3 4 #define MAXN 200005 5 #define lson l, mid, rt<<1 6 #define rson mid+1, r, rt<<1|1 7 #define mymax(x, y) (x>y) ? x:y 8 9 int nums[MAXN<<2]; 10 int h, w; 11 12