題目描述

Ivan unexpectedly saw a present from one of his previous birthdays. It is array of n numbers from 1 to 200. Array is old and some numbers are hard to read. Ivan remembers that for all elements at least one of its neighbours ls not less than it, more formally:
a1≤a2,
an≤an?1?and
ai≤max(ai?1,ai+1) for all i from 2 to n?1.
Ivan does not remember the array and asks to find the number of ways to restore it. Restored elements also should be integers from 1 to 200. Since the number of ways can be big, print it modulo 998244353.

輸入

First line of input contains one integer n (2≤n≤105) — size of the array.

Second line of input contains n integers ai — elements of array. Either ai=?1 or 1≤ai≤200. ai=?1 means that i-th element can't be read.

輸出

Print number of ways to restore the array modulo 998244353.

樣例輸入

3 1 -1 2

樣例輸出

1
題意 構造一個長度為n的序列，有些位置是-1，可以填1-200的數字，要使得每個位置都比它左右兩側的最大值小，求方案數思路 dp f[i][j][0/1/2]表示到第i位，當前數為j，從i-1到i是上升/相等/下降的方案數 顯然 f[i][j][0]=f[i-1][k][0]+f[i-1][k][1]+f[i-1][k][2]; k<j; f[i][j][1]=f[i-1][k][0]+f[i-1][k][1]+f[i-1][k][2]; k=j f[i][j][2]=f[i-1][k][1]+f[i-1][k][2]; #include <bits/stdc++.h> #define ll long long using namespace std; const int P=998244353; const int N=1e5+10; ll f[N][205][3]; ll sum[2][205][3]; int a[N]; int n; int main(){scanf("%d",&n);for (int i=1;i<=n;i++) scanf("%d",&a[i]);if (a[1]==-1){for (int i=1;i<=200;i++) f[1][i][0]=1;} else f[1][a[1]][0]=1;for(int i = 1; i <= 200; i++) sum[0][i][0] = (sum[0][i-1][0] + f[1][i][0])%P;for (int i=2;i<=n;i++) {//sum[!(i&1)][0][0] = sum[!(i&1)][0][1] = sum[!(i&1)][0][2] = 0;for (int j=1;j<=200;j++) {if (a[i]==-1 || a[i]==j){//f[i][j][0]=f[i-1][k][0]+f[i-1][k][1]+f[i-1][k][2]; k<j;f[i][j][0]=((sum[i&1][j-1][0]+sum[i&1][j-1][1])%P+sum[i&1][j-1][2])%P;//f[i][j][1]=f[i-1][k][0]+f[i-1][k][1]+f[i-1][k][2]; k=jf[i][j][1]=(f[i-1][j][0]+f[i-1][j][1]+f[i-1][j][2])%P;//f[i][j][2]=(f[i][j][2]+f[i-1][k][1]+f[i-1][k][2])%p;f[i][j][2]=((sum[i&1][200][1] - sum[i&1][j][1] +P)%P + (sum[i&1][200][2] - sum[i&1][j][2]+P)%P)%P;}sum[!(i&1)][j][0] = (sum[!(i&1)][j-1][0] + f[i][j][0])%P;sum[!(i&1)][j][1] = (sum[!(i&1)][j-1][1] + f[i][j][1])%P;sum[!(i&1)][j][2] = (sum[!(i&1)][j-1][2] + f[i][j][2])%P;}}// cout<<f[1][a[1]][0]<<' '<<f[1][a[1]][1]<<' '<<f[1][a[1]][2]<<endl;ll ans=0;if (a[n]==-1){for (int i=1;i<=200;i++){// printf("f[3][%d][0]=%lld,f[3][%d][1]=%lld,f[3][%d][2]=%lld\n",i,f[3][i][0],i,f[3][i][1],i,f[3][i][2]);ans=(ans+f[n][i][1]+f[n][i][2])%P;}} else ans=(f[n][a[n]][1]+f[n][a[n]][2])%P;printf("%lld\n",ans);return 0; } View Code

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k>j 枚舉k的話是200*200*n，所以要前綴和優化……但可能寫的過于詭異

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轉載于:https://www.cnblogs.com/tetew/p/11317677.html

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