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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#751440 | #5117. Find Maximum | HaijieTan | WA | 0ms | 3544kb | C++23 | 2.1kb | 2024-11-15 18:43:55 | 2024-11-15 18:43:56 |
Judging History
answer
#include<iostream>
#include<vector>
#include<bitset>
#include<algorithm>
#include<cmath>
#include<queue>
#include<numeric>//iota -- iota(all(a), x) a从头到尾从x开始递增
#include<cstring>
#include<ctime>
#include<array>
#include<list>
#include<random>
#include<tuple>
#include<deque>
#include<map>
#include<stack>
#include<set>
using namespace std;
template <typename Dtype>
unsigned int __builtin_popcount(Dtype u)
{
u = (u & 0x55555555) + ((u >> 1) & 0x55555555);
u = (u & 0x33333333) + ((u >> 2) & 0x33333333);
u = (u & 0x0F0F0F0F) + ((u >> 4) & 0x0F0F0F0F);
u = (u & 0x00FF00FF) + ((u >> 8) & 0x00FF00FF);
u = (u & 0x0000FFFF) + ((u >> 16) & 0x0000FFFF);
return u;
}//wishchin!!!
#define IOS ios::sync_with_stdio(false);cin.tie(0), cout.tie(0)
#define int long long
#define all(x) x.begin(), x.end()
#define pb push_back
#define pf push_front
#define lowbit(x) (x & (-x))
const int xy[][2] = { 0,1, 0, -1, 1, 0, -1, 0 };
#define mem(a, x) memset(a, x, sizeof a)
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<char, char> PCC;
typedef pair<int, char> PIC;
typedef pair<char, int> PCI;
#define x first
#define y second
#define NO cout << "NO\n"
#define YES cout << "YES\n"
inline int read();
const int INF = 0x3f3f3f3f, N = 1e6 + 10, MOD = 1e9 + 7, M = 1e4 + 10, K = 131;
const double PI = acos(-1.0);
void solve() {
int n; cin >> n;
vector<int>a(n + 2), dp(n + 2, 0);
dp[n - 1] = 1;
for (int i = 0; i < n; i++)cin >> a[i];
auto get = [&](int pos) -> int {
if (pos > n)return n + 1;
if (pos == n)return 0;
return dp[pos];
};
for (int i = n - 2; ~i; i--)
dp[i] = min(dp[i + 1] + 1, get(i + a[i] + 1));
cout << dp[0] << endl;
}
signed main() {
IOS;
int _ = 1;
cin >> _;
while (_--)solve();
return 0;
}
inline int read()
{
int x = 0, f = 1; char ch = getchar();
while (ch < '0' || ch>'9') { if (ch == '-') f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
return x * f;
}
/* /\_/\
* (= ._.)
* / > \>
*/
Details
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Test #1:
score: 0
Wrong Answer
time: 0ms
memory: 3544kb
input:
10 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5
output:
1 1 1 1 2 1 3 1 1 1
result:
wrong answer 1st numbers differ - expected: '3', found: '1'