阿力的第二場難題

A:

本意是當簽到題，但是可能低估了這題的難度

在線標記處理二分查找

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000005;
int n, w[maxn];
bool vis[maxn];
set<int> a;
set<int>::iterator it;
inline int read() {
    int ret = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')
            f = -f;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') ret = (ret << 3) + (ret << 1) + ch - '0', ch = getchar();
    return ret * f;
}
int main() {
    n = read();
    for (int i = 1; i <= n; i++) {
        char ch = getchar();
        int x;
        while (ch != 'O' && ch != 'I' && ch != '?') ch = getchar();
        if (ch == '?')
            a.insert(i);
        else if (ch == 'I') {
            x = read();
            if (vis[x]) {
                it = a.upper_bound(w[x]);
                if (it == a.end() || (*it > i))
                    return printf("%d\n", i), 0;
                else
                    a.erase(it);
            }
            vis[x] = 1;
            w[x] = i;
        } else {
            x = read();
            if (!vis[x]) {
                it = a.upper_bound(w[x]);
                if (it == a.end() || (*it > i))
                    return printf("%d\n", i), 0;
                else
                    a.erase(it);
            }
            vis[x] = 0, w[x] = i;
        }
    }
    printf("-1\n");
    return 0;
}

B題

本意通過素數篩求，但是有一組測試數據太大沒放上去，導致別的做法可以卡過去，因爲測評機的原因，我開了3s，降低了難度係數

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e8 + 110, mod = 100000007;
int n, pri[10000000], tot = 0;
long long ans = 1;
bool hs[maxn];
void shai() {
    hs[1] = 1;
    for (int i = 2; i <= n; ++i) {
        if (!hs[i]) {
            pri[++tot] = i;
            for (long long k = i; k <= n; k *= i) ans = (ans * i) % mod;
        }
        for (int j = 1; j <= tot && i * pri[j] <= n; ++j) {
            hs[i * pri[j]] = 1;
            if (!(i % pri[j]))
                break;
        }
    }
}
int main() {
    scanf("%d", &n);
    shai();
    printf("%d\n", ans);
    return 0;
}

C:

邏輯題，其實沒啥東西。。

long long a[1000010], b[1000010], v, n;
int main() {
    scanf("%d", &n);
    b[0] = 1;
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &a[i]);
        (v += a[i]) %= n;
        if (b[v]) {
            for (int j = b[v]; j <= i; j++) printf("%d %d\n", j, a[j]);
            return 0;
        }
        b[v] = i + 1;
    }
    return 0;
}

D：

min_25板子題，其實我一開始沒考慮板子，後來cc提到的

#include <bits/stdc++.h>
#define ri register int
using namespace std;
const int N = 1e6 + 5;
typedef long long ll;
ll f1[N], f2[N], n;
int lim;
int main() {
    cin >> n;
    lim = sqrt(n);
    for (ri i = 1; i <= lim; ++i) f1[i] = i - 1, f2[i] = n / i - 1;
    for (ri p = 2; p <= lim; ++p) {
        if (f1[p] == f1[p - 1])
            continue;
        for (ri i = 1; i <= lim / p; ++i) f2[i] -= f2[i * p] - f1[p - 1];
        for (ri i = lim / p + 1; (ll)i <= n / (ll)p / (ll)p && i <= lim; ++i)
            f2[i] -= f1[n / i / p] - f1[p - 1];
        for (ri i = lim; i >= (ll)p * p; --i) f1[i] -= f1[i / p] - f1[p - 1];
    }
    cout << f2[1];
    return 0;
}

E:斐波那契矩陣快速冪

這是我的標程序，可以記下來當板子，但是不是很好理解，建議別看。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
const int mod=1e9+7;

pll operator *(const pll &a,const pll &b)
{
    return make_pair((a.first*b.first+a.second*b.second)%mod,(a.first*b.second+a.second*b.first+a.second*b.second)%mod);
}
ll F(ll y)
{
    pll ans=make_pair(1,0),x=make_pair(0,1);
    while(y)
    {
        if(y&1) ans=ans*x;
        x=x*x;
        y/=2;
    }
    return ans.first+ans.second;
}

int main()
{
    ll n;
    cin>>n;
    cout<<F(n-1)*F(n)%mod;
    return 0;
}

這是常規做法

學這個吧

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
LL n,p;
struct node
{
    LL m[3][3];
    node(){
        memset(m,0,sizeof(m));
    }
    friend node operator *(node x,node y){
        node t;
        for(int i=1;i<=2;i++){
            for(int j=1;j<=2;j++){
                t.m[i][j]=0;
            }
        }
        for(int i=1;i<=2;i++){
            for(int j=1;j<=2;j++){
                for(int k=1;k<=2;k++){
                    t.m[i][j]+=((x.m[i][k]%mod)*(y.m[k][j]%mod))%mod;
                }
            }
        }
        return t;
    }
}a,ans;
node binary(node x,LL y)
{
    while(y){
        if(y&1) ans=ans*x;
        x=x*x;
        y>>=1;
    }
    return ans;
    
}
void Init()
{
    a.m[1][1]=1;
    a.m[1][2]=1;
    a.m[2][1]=1;
    a.m[2][2]=0;
    
    ans.m[1][1]=1;
    ans.m[1][2]=1;
}
int main()
{
    Init();
    LL n;scanf("%lld",&n);
    n++;
    if(n<=2) printf("1\n");
    else{
        node t=binary(a,n-2);
        printf("%lld\n",((t.m[1][1]%mod)*(t.m[1][2]%mod))%mod);
    } 
    return 0;
}