HDU 5719 BestCoder 2nd Anniversary Arrange （DP）

Arrange

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 558    Accepted Submission(s): 198

Problem Description

Accidentally, Cupid, god of desire has hurt himself with his own dart and fallen in love with Psyche.

This has drawn the fury of his mother, Venus. The goddess then throws before Psyche a great mass of mixed crops.

There are n heaps of crops in total, numbered from 1 to n.

Psyche needs to arrange them in a certain order, assume crops on the i-th position is Ai.

She is given some information about the final order of the crops:

1. the minimum value of A1,A2,...,Ai is Bi.

2. the maximum value of A1,A2,...,Ai is Ci.

She wants to know the number of valid permutations. As this number can be large, output it modulo998244353.

Note that if there is no valid permutation, the answer is 0.

 

Input

The first line of input contains an integerT(1≤T≤15), which denotes the number of testcases.

For each test case, the first line of input contains single integer n(1≤n≤105).

The second line contains n integers, the i-th integer denotes Bi(1≤Bi≤n).

The third line contains n integers, the i-th integer denotes Ci(1≤Ci≤n).

Output

For each testcase, print the number of valid permutations modulo998244353.

Sample Input

232 1 12 2 355 4 3 2 11 2 3 4 5

Sample Output

10
Hint
In the first example, there is only one valid permutation (2,1,3) .In the second example, it is obvious that there is no valid permutation.
 

Source

BestCoder 2nd Anniversary

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题目大意: 求出合法的1~n的全排列的种数。

         他会给你两行数。

         第一行表示的Ai 表示这个排列的前 i 个的最小值为Ai。

         第二行表示的B i表示这个排列的前 i 个的最大值为Bi。、

         要你求出符合这些条件的全排列的种数。

题解：首先排除掉5种不可能的情况,1.b[i]>b[i-1] 2,c[i]<c[i-1] 3,b[i]>c[i] 4.c[1]!=b[1] 5.b[i],c[i] < 1 || > n ,
           然后递推,如果当前产生的新的 b[i]或者 c[i] 那么dp[i] = dp[i-1] ，
           如果当前 b[i-1] = b[i] && c[i-1] = c[i] ，那么我们可以在 [b[i],c[i]]中任选一个数,
           但是由于这些数是互不相同的,所以每次我们的选的数都会逐渐变少。
           用num计算一下当前已经选了多少种,减掉之后答案即为 dp[i] = dp[i-1]*(c[i]-b[i]+1-num)。

AC代码：

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm> using namespace std;typedef long long LL;const int mod =998244353;const int N=100005;int b[N];int c[N];LL dp[N];int main(){    int t;    scanf("%d",&t);    while(t--)    {    int n;    scanf("%d",&n);    int Min=999999;    int Max=-1;    int flag=1;    for(int i=1;i<=n;i++)    {    scanf("%d",&b[i]);    if(b[i]>Min) flag=0;    if(b[i]<1||b[i]>n) flag=0;    Min=min(Min,b[i]);}for(int i=1;i<=n;i++){scanf("%d",&c[i]);if(c[i]<Max)flag=0;if(c[i]<1||c[i]>n) flag=0;if(c[i]<b[i]) flag=0;Max=max(Max,c[i]);}if(!flag||c[1]!=b[1]) puts("0"); //前 1个的最大最小值一定相等 ，都是这个数嘛 else{memset(dp,0,sizeof(dp));dp[1]=1;int num=1;for(int i=2;i<=n;i++) //dp核心 {if(c[i]==c[i-1]&&b[i]==b[i-1]&&b[i]!=c[i]){dp[i]=dp[i-1]*(c[i]-b[i]+1-num)%mod;}else if( b[i]<b[i-1]&&c[i-1]==c[i] || b[i]==b[i-1]&&c[i-1]<c[i] ){//dp[i] = dp[i-1]+1;dp[i]=dp[i-1];}num++; //已经选了多少种}  printf("%I64d\n",dp[n]);}}    return 0;}