有一个含有 n 个数字的序列，每个数字的大小是不超过 200 的正整数，同时这个序列满足以下条件：
但是很不幸的是，在序列保存的过程中，有些数字丢失了，请你根据上述条件，计算可能有多少种不同的序列可以满足以上条件。
数据范围： 
， 序列中的数字满足 
， 数字为 0 时表示丢失
[编程题]还原
- 热度指数：2280 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 256M，其他语言512M
- 算法知识视频讲解
输入描述:
输入第一行是一个n，表示这个序列的长度。
输入第二行有n个非负整数，中间用空格隔开，如果数字为0，说明这个数字丢失了，其他数字则都在1-200之间。
输出描述:
输出仅包含一个整数，即方案数对998244353取模的结果。
示例1
输入
3 2 0 1
输出
1
说明
第二个数要求大于等于第一个数，则第二个数大于等于 2，且根据第二个条件，必须大于等于最后一个数，则大于等于 1 ，根据第三个条件，必须小于等于左边和右边的数的最大值，则小于等于 2 ，大于等于 2 ，所以序列只可为 2 2 1
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4+3;
const int M = 998244353;
int main(){
int n;
scanf("%d", &n);
int a[n+2];
for(int i=1;i<=n;i++)
scanf("%d", &a[i]);
a[0] = a[n+1] = 1;
long dp[n+2][203][3], s1[203], s2[203];
dp[0][1][1] = 1;
for(int i=1;i<=200;i++)
s1[i] = s2[i] = 1;
for(int i=1;i<=n+1;i++){
int l=1, r=200;
if(a[i])
l = r = a[i];
for(int j=l;j<=r;j++){
dp[i][j][0] = (s2[200]-s2[j])%M;
dp[i][j][1] = (dp[i-1][j][0]+dp[i-1][j][1]+dp[i-1][j][2])%M;
dp[i][j][2] = s1[j-1]%M;
}
for(int j=1;j<=200;j++){
s1[j] = s1[j-1] + dp[i][j][0] + dp[i][j][1] + dp[i][j][2];
s2[j] = s2[j-1] + dp[i][j][0] + dp[i][j][1];
}
}
printf("%ld\n", (dp[n+1][1][0]+dp[n+1][1][1])%M);
return 0;
}
发表于 2020-10-10 01:02:41
回复(0)
#include <bits/stdc++.h>
using namespace std;
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
#define LOWBIT(x) ((x)&(-x))
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))
#define MP make_pair
#define PB push_back
#define ft first
#define sd second
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
in >> p.first >> p.second;
return in;
}
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
for (auto &x: v)
in >> x;
return in;
}
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
out << "[" << p.first << ", " << p.second << "]" << "\n";
return out;
}
inline int gc(){
static const int BUF = 1e7;
static char buf[BUF], *bg = buf + BUF, *ed = bg;
if(bg == ed) fread(bg = buf, 1, BUF, stdin);
return *bg++;
}
inline int ri(){
int x = 0, f = 1, c = gc();
for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
return x*f;
}
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< string, int > PSI;
typedef set< int > SI;
typedef vector< int > VI;
typedef vector< PII > VPII;
typedef map< int, int > MII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
const double EPS = 1e-10;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 998244353;
const int maxN = 1e4 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;
int n, a[maxN];
// dp[i][j][0/1/2] 代表以a[i]为结尾，a[i] = j时的序列种数。
// 第三维代表a[i-1]和a[i]的大小关系(>, ==, <)。
LL dp[maxN][207][3];
LL preSum1[207], preSum2[207];
int main(){
INIT();
cin >> n;
For(i, 1, n) cin >> a[i];
a[0] = a[n + 1] = 1;// a[-1] = 1 // 一共3个哨兵
// 预处理
dp[0][1][1] = 1;
For(i, 1, 200) preSum1[i] = preSum2[i] = 1;
For(i, 1, n + 1) {
int s = 1, t = 200;
if(a[i]) s = t = a[i];
For(j, s, t) {
dp[i][j][0] = (preSum2[200] - preSum2[j]) % mod;
dp[i][j][1] = (dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2]) % mod;
dp[i][j][2] = preSum1[j - 1] % mod;
}
// 更新前缀和
For(j, 1, 200) {
preSum1[j] = preSum1[j - 1] + dp[i][j][0] + dp[i][j][1] + dp[i][j][2];
preSum2[j] = preSum2[j - 1] + dp[i][j][0] + dp[i][j][1];
}
}
cout << (dp[n + 1][1][0]+dp[n + 1][1][1]) % mod << endl;
return 0;
}
发表于 2021-10-08 16:11:57
回复(1)
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#include <iostream>
#include <vector>
#include <algorithm>
#define int long long
using namespace std;
const int MAXN = 1e4 + 10;
const int mod = 998244353;
int dp[MAXN][205][2];
int nums[MAXN];
int sum1[205];
int sum2[205];
int sum3[205];
int sum4[205];
int n;
signed main()
{
cin >> n;
for (int i = 1; i <= n; i++) cin >> nums[i];
for (int j = 0; j <= 200; j++)
{
dp[0][j][1] = 1;
dp[0][j][0] = 0;
sum3[j] = j == 0 ? dp[0][j][0] : (sum3[j - 1] + dp[0][j][0]);
sum4[j] = j == 0 ? dp[0][j][1] : (sum4[j - 1] + dp[0][j][1]);
}
for (int i = 1; i <= n; i++)
{
for (int j = 0; j <= 200; j++)
{
sum1[j] = sum3[j];
sum2[j] = sum4[j];
}
for (int j = 0; j <= 200; j++)
{
for (int s = 0; s <= 1; s++)
{
int ans = 0;
if (nums[i] != 0)
{
if (nums[i] >= j || s == 1)
ans = dp[i - 1][nums[i]][nums[i] > j ? 0 : 1];
}
else
{
ans += sum1[200] - sum1[j];
if (j != 0) ans += dp[i - 1][j][1];
if (s == 1 && j != 0) ans += sum2[j - 1] - sum2[0];
}
ans %= mod;
dp[i][j][s] = ans;
}
sum3[j] = j == 0 ? dp[i][j][0] : (sum3[j - 1] + dp[i][j][0]);
sum4[j] = j == 0 ? dp[i][j][1] : (sum4[j - 1] + dp[i][j][1]);
}
}
cout << dp[n][0][1];
return 0;
}
发表于 2025-02-06 10:01:42
回复(0)
#include <vector>
#include <math.h>
#include <math.h>
#include <iostream>
using namespace std;
void test02()
{
vector<int> NUM;
int n;
cin >> n;
int k = 0;
while (k < n)
{
k++;
int tmp;
cin >> tmp;
if (tmp == 13)
{
break;
}
NUM.push_back(tmp);
}
vector<int> output;
for (int i = 0; i < n - 1; i++)
{
if (NUM[i] == 0)
{
int may = 0;
for (may = 1; may <= 200; may++)
{
if (i == 0)
{
if (may <= NUM[1]);
output.push_back(may);// cout << "第" << i + 1 << "位置" << NUM[i] << endl;
}
else if (i == 1)
{
if (may >= NUM[0] && may <= max(NUM[i - 1], NUM[i + 1]) && may >= NUM[i + 1])
output.push_back(may);//cout << "第" << i + 1 << "位置" << may << endl;
}
else if (i == n)
{
if (may >= NUM[i + 1]);
output.push_back(may);//cout << NUM[i] << endl;
}
else
{
if (may >= NUM[i + 1] && may > max(NUM[i - 1], NUM[i + 1]))//判断是否满足条件
{
continue;
}
else {
output.push_back(may);//cout << "第" << i + 1 << "位置" << may << endl;
}
}
}
}
}
for (int i=0;i<output.size(); i++)
cout << output[i] << endl;
}
int main() {
test02();
return 0;
}
{
vector<int> NUM;
int n;
cin >> n;
int k = 0;
while (k < n)
{
k++;
int tmp;
cin >> tmp;
if (tmp == 13)
{
break;
}
NUM.push_back(tmp);
}
vector<int> output;
for (int i = 0; i < n - 1; i++)
{
if (NUM[i] == 0)
{
int may = 0;
for (may = 1; may <= 200; may++)
{
if (i == 0)
{
if (may <= NUM[1]);
output.push_back(may);// cout << "第" << i + 1 << "位置" << NUM[i] << endl;
}
else if (i == 1)
{
if (may >= NUM[0] && may <= max(NUM[i - 1], NUM[i + 1]) && may >= NUM[i + 1])
output.push_back(may);//cout << "第" << i + 1 << "位置" << may << endl;
}
else if (i == n)
{
if (may >= NUM[i + 1]);
output.push_back(may);//cout << NUM[i] << endl;
}
else
{
if (may >= NUM[i + 1] && may > max(NUM[i - 1], NUM[i + 1]))//判断是否满足条件
{
continue;
}
else {
output.push_back(may);//cout << "第" << i + 1 << "位置" << may << endl;
}
}
}
}
}
for (int i=0;i<output.size(); i++)
cout << output[i] << endl;
}
int main() {
test02();
return 0;
}
发表于 2022-09-23 19:28:14
回复(0)
//回溯法 想想都应该是超时的 错误提示时间复杂度过大
#include<iostream>
(720)#include<vector>
#include<algorithm>
using namespace std;
const long long ModNum = 998244353;
long long ans = 0;
void BackTrack(vector<int> &v, vector<int> temp, int i) {
int maxNum;
if (i == v.size()) {
ans++;
ans %= 998244353;
return;
}
if (v[i] != 0) {
if (i == 0) {
temp.push_back(v[i]);
BackTrack(v, temp, i + 1);
temp.pop_back();
}
else if (i == 1) {
if (v[i] >= temp.back()) {
temp.push_back(v[i]);
BackTrack(v, temp, i + 1);
temp.pop_back();
}
}
else {
maxNum = max(temp[temp.size() - 2], v[i]);
if (temp[temp.size() - 1] <= maxNum && i != v.size() - 1) {
temp.push_back(v[i]);
BackTrack(v, temp, i + 1);
temp.pop_back();
}
else if (temp[temp.size() - 1] <= maxNum && i == v.size() - 1 && v[i] <= temp.back()){
temp.push_back(v[i]);
BackTrack(v, temp, i + 1);
temp.pop_back();
}
}
}
else {
for (int j = 1; j <= 200; j++) {
if (i == 0) {
temp.push_back(j);
BackTrack(v, temp, i + 1);
temp.pop_back();
}
else if (i == 1) {
if (j >= temp.back()) {
temp.push_back(j);
BackTrack(v, temp, i + 1);
temp.pop_back();
}
}
else {
maxNum = max(temp[temp.size() - 2], j);
if (temp[temp.size() - 1] <= maxNum && i != v.size()-1) {
temp.push_back(j);
BackTrack(v, temp, i + 1);
temp.pop_back();
}
else if (temp[temp.size() - 1] <= maxNum && i == v.size() - 1 && j <= temp.back()) {
temp.push_back(j);
BackTrack(v, temp, i + 1);
temp.pop_back();
}
}
}
}
return;
}
int main(void) {
int n;
cin >> n;
vector<int> v;
int num;
for (int i = 0; i < n; i++) {
cin >> num;
v.push_back(num);
}
vector<int> temp;
BackTrack(v, temp, 0);
cout << ans << endl;
return 0;
}
发表于 2020-05-09 18:18:00
回复(0)
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long mod = 998244353;
int N = sc.nextInt();
int[] arr = new int[N + 10];
for (int i = 1; i <= N; i++) {
arr[i] = sc.nextInt();
}
long[][][] res = new long[N + 10][210][3];
long[] s1 = new long[210];
long[] s2 = new long[210];
//初始化
for (int i = (arr[1] == 0 ? 1 : arr[1]); i <= (arr[1] == 0 ? 200 : arr[1]); i++) {
for (int j = (arr[2] == 0 ? 1 : arr[2]); j <= (arr[2] == 0 ? 200 : arr[2]); j++) {
if (i == j) res[2][j][1]++;
else if (i < j) res[2][j][2]++;
}
}
for (int j = 1; j <= 200; j++) {
s1[j] = s1[j - 1] + res[2][j][0] + res[2][j][1] + res[2][j][2];
s2[j] = s2[j - 1] + res[2][j][0] + res[2][j][1];
}
//更新结构
for (int i = 3; i <= N; i++) {
for (int j = (arr[i] == 0 ? 1 : arr[i]); j <= (arr[i] == 0 ? 200 : arr[i]); j++) {
res[i][j][0] = (s2[200] - s2[j]) % mod;
res[i][j][1] = (res[i - 1][j][0] + res[i - 1][j][1] + res[i - 1][j][2]) % mod;
res[i][j][2] = s1[j - 1] % mod;
}
for (int j = 1; j <= 200; j++) {
s1[j] = s1[j - 1] + res[i][j][0] + res[i][j][1] + res[i][j][2];
s2[j] = s2[j - 1] + res[i][j][0] + res[i][j][1];
}
}
long ans = 0;
for (int j = (arr[N] == 0 ? 1 : arr[N]); j <= (arr[N] == 0 ? 200 : arr[N]); j++) {
ans = (ans + res[N][j][0] + res[N][j][1]) % mod;
}
System.out.println(ans);
}
}
https://www.acwing.com/solution/acwing/content/1873/
发表于 2020-04-16 23:23:56
回复(0)
问题信息
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2023-02-08 23:50:43
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