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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#424882#7109. Traveling on the AxisPCTprobabilityAC ✓1ms3992kbC++144.5kb2024-05-29 19:27:222024-05-29 19:27:23

Judging History

你现在查看的是最新测评结果

  • [2024-05-29 19:27:23]
  • 评测
  • 测评结果:AC
  • 用时:1ms
  • 内存:3992kb
  • [2024-05-29 19:27:22]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
using ll = long long;
using ld = long double;
using ull = unsigned long long;
#define endl "\n"
typedef pair<int, int> Pii;
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REP3(i, m, n) for (int i = (m); (i) < int(n); ++ (i))
#define rep(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define ALL(x) begin(x), end(x)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(s) (s).begin(),(s).end()
#define drep2(i, m, n) for (int i = (m)-1; i >= (n); --i)
#define drep(i, n) drep2(i, n, 0)
#define rever(vec) reverse(vec.begin(), vec.end())
#define sor(vec) sort(vec.begin(), vec.end())
#define fi first
#define FOR_(n) for (ll _ = 0; (_) < (ll)(n); ++(_))
#define FOR(i, n) for (ll i = 0; (i) < (ll)(n); ++(i))
#define se second
#define pb push_back
#define P pair<ll,ll>
#define PQminll priority_queue<ll, vector<ll>, greater<ll>>
#define PQmaxll priority_queue<ll,vector<ll>,less<ll>>
#define PQminP priority_queue<P, vector<P>, greater<P>>
#define PQmaxP priority_queue<P,vector<P>,less<P>>
#define NP next_permutation
#define die(a) {cout<<a<<endl;return 0;}
#define dier(a) {return a;}
//const ll mod = 1000000009;
const ll mod = 998244353;
//const ll mod = 1000000007;
const ll inf = 4100000000000000000ll;
const ld eps = ld(0.00000000001);
static const long double pi = 3.141592653589793;
template<class T>void vcin(vector<T> &n){for(int i=0;i<int(n.size());i++) cin>>n[i];}
template<class T,class K>void vcin(vector<T> &n,vector<K> &m){for(int i=0;i<int(n.size());i++) cin>>n[i]>>m[i];}
template<class T>void vcout(vector<T> &n){for(int i=0;i<int(n.size());i++){cout<<n[i]<<" ";}cout<<endl;}
template<class T>void vcin(vector<vector<T>> &n){for(int i=0;i<int(n.size());i++){for(int j=0;j<int(n[i].size());j++){cin>>n[i][j];}}}
template<class T>void vcout(vector<vector<T>> &n){for(int i=0;i<int(n.size());i++){for(int j=0;j<int(n[i].size());j++){cout<<n[i][j]<<" ";}cout<<endl;}cout<<endl;}
void yes(bool a){cout<<(a?"yes":"no")<<endl;}
void YES(bool a){cout<<(a?"YES":"NO")<<endl;}
void Yes(bool a){cout<<(a?"Yes":"No")<<endl;}
void possible(bool a){ cout<<(a?"possible":"impossible")<<endl; }
void Possible(bool a){ cout<<(a?"Possible":"Impossible")<<endl; }
void POSSIBLE(bool a){ cout<<(a?"POSSIBLE":"IMPOSSIBLE")<<endl; }
#define FOR_R(i, n) for (ll i = (ll)(n)-1; (i) >= 0; --(i))
template<class T>auto min(const T& a){ return *min_element(all(a)); }
template<class T>auto max(const T& a){ return *max_element(all(a)); }
template<class T,class F>void print(pair<T,F> a){cout<<a.fi<<" "<<a.se<<endl;}
template<class T>bool chmax(T &a,const T b) { if (a<b) { a=b; return 1; } return 0;}
template<class T>bool chmin(T &a,const T b) { if (b<a) { a=b; return 1; } return 0;}
template<class T> void ifmin(T t,T u){if(t>u){cout<<-1<<endl;}else{cout<<t<<endl;}}
template<class T> void ifmax(T t,T u){if(t>u){cout<<-1<<endl;}else{cout<<t<<endl;}}
ll fastgcd(ll u,ll v){ll shl=0;while(u&&v&&u!=v){bool eu=!(u&1);bool ev=!(v&1);if(eu&&ev){++shl;u>>=1;v>>=1;}else if(eu&&!ev){u>>=1;}else if(!eu&&ev){v>>=1;}else if(u>=v){u=(u-v)>>1;}else{ll tmp=u;u=(v-u)>>1;v=tmp;}}return !u?v<<shl:u<<shl;}
ll modPow(ll a, ll n, ll mod) { if(mod==1) return 0;ll ret = 1; ll p = a % mod; while (n) { if (n & 1) ret = ret * p % mod; p = p * p % mod; n >>= 1; } return ret; }
vector<ll> divisor(ll x){ vector<ll> ans; for(ll i = 1; i * i <= x; i++){ if(x % i == 0) {ans.push_back(i); if(i*i!=x){ ans.push_back(x / ans[i]);}}}sor(ans); return ans; }
ll pop(ll x){return __builtin_popcountll(x);}
ll poplong(ll x){ll y=-1;while(x){x/=2;y++;}return y;}
P hyou(P a){ll x=fastgcd(abs(a.fi),abs(a.se));a.fi/=x;a.se/=x;if(a.se<0){a.fi*=-1;a.se*=-1;}return a;}
P Pplus(P a,P b){ return hyou({a.fi*b.se+b.fi*a.se,a.se*b.se});}
P Ptimes(P a,ll b){ return hyou({a.fi*b,a.se});}
P Ptimes(P a,P b){ return hyou({a.fi*b.fi,a.se*b.se});}
P Pminus(P a,P b){ return hyou({a.fi*b.se-b.fi*a.se,a.se*b.se});}
P Pgyaku(P a){ return hyou({a.se,a.fi});}

void cincout(){
  ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
  cout<< fixed << setprecision(15);
}
void solve(){
  string s;
  cin>>s;
  ll n=s.size();
  ll ans=0;
  for(ll i=0;i+1<n;i++){
    if(s[i]==s[i+1]) ans+=(i+1)*(n-i-1);
  }
  for(int i=0;i<n;i++) if(s[i]=='0') ans+=n-i;
  ans+=n*(n+1)*(n+2)/6;
  cout<<ans<<endl;
}
int main(){
  cincout();
  ll t;
  cin>>t;
  while(t--) solve();
}

这程序好像有点Bug,我给组数据试试?

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 3652kb

input:

3
101
011
11010

output:

12
15
43

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 0ms
memory: 3992kb

input:

6107
1010101
010110100110101
1010
1010101010010101010
101011
0101101011010101010
0101101011
11011010101
010
1011010
10110101010101010100
010101010110101
10101010101011
0101010101010101011
00101010011000
1010101010010110110
01010101001010101010
101010101010101
100100101010101010
01
011
0101010100101
...

output:

96
889
24
1515
69
1567
279
345
14
106
1702
791
621
1447
764
1615
1755
736
1333
6
15
542
44
1689
1515
140
833
497
596
24
1640
694
462
30
425
14
1041
1446
96
504
124
75
560
970
771
945
6
1
321
137
786
720
206
769
46
103
225
74
554
2
100
529
260
207
197
2
197
1041
140
857
207
1
74
1604
41
343
1041
14
1...

result:

ok 6107 lines

Extra Test:

score: 0
Extra Test Passed