Ad Infinitum 8

比赛链接

A题

如果当前数是１，那么后面无论是几都会加１或者当后面数是１的时候加２，所以记录一下后面的数中１的个数（加２）即可。

如果当前数是２，那么只有当后面的数为１或者为２时才可以加１，所以再记录一下后面数中２的个数即可。

#include <map>
#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int a[200050], t, n, num1[200050], num2[200050];

int main() {
    ios::sync_with_stdio(false);
    cin >> t;
    while (t--) {
        cin >> n;
        for (int i = 1; i <= n; i++) cin >> a[i];
        memset(num1, 0, sizeof(num1));
        memset(num2, 0, sizeof(num2));
        for (int i = n; i >= 1; i--) {
            num1[i] = num1[i + 1];
            num2[i] = num2[i + 1];
            if (a[i] == 1) num1[i]++;
            else if (a[i] == 2) num2[i]++;
        }
        long long res = 0;
        for (int i = 1; i <= n; i++) {
            if (a[i] == 1) {
                res += n - i + num1[i + 1];
            } else if (a[i] == 2) {
                res += num1[i + 1] + num2[i + 1];
            } else {
                res += num1[i + 1];
            }
        }
        cout << res << endl;
    }
    return 0;
}

B题

这题需要敏锐的洞察力啊，起码对我来说是这样！

手动向后推两项就可以发现，当Ｎ为２时，Ｓ = (a1 + 1) * (a2 + 1) - 1; 

当N为３时，S = (a1 + 1) * (a2 + 1) * (a3 + 1) - 1

可以发现，一个序列的S值和其顺序无关，只和数有关。

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

const int MOD = 1e9 + 7;
const int MAX_N = 10050;
const int MAX_V = 1000050; 
typedef long long LL;
#define rep(i, n) for (int i = (1); i <= (n); i++)

int main() {
    ios::sync_with_stdio(false);
    int N;
    LL res = 0;
    cin >> N;
    rep (i, N) {
        int x;
        cin >> x;
        res = (res + x + res * x) % MOD;
    }
    cout << res << endl;
    return 0;
}

C题

很裸的求逆，快速幂。

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int extgcd(int a, int b, int &x, int &y) {
    int d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}

int mod_inverse(int a, int m) {
    int x, y;
    extgcd(a, m, x, y);
    return (m + x % m) % m;
}

int mod_pow(int a, int b, int m) {
    long long res = 1;
    while (b) {
        if (b & 1) res = res * a % m;
        a = ((long long)a * a) % m;
        b >>= 1;
    }
    return res;
}

int main() {
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
        int a, b, x;
        cin >> a >> b >> x;
        if (b < 0) {
            a = mod_inverse(a, x);
            b = -b;
        }
        cout << mod_pow(a, b, x) << endl;
    }
    return 0;
}

D题

容斥原理模板题

#include <map>
#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
const int MAX_N = 20;
int N;
int num[MAX_N];
LL n, ans;
LL gcd(LL a, LL b){
    LL r;
    while( b ){
        r = a%b;
        a = b;
        b = r;
    }
    return a;
}

void dfs( int id, LL Lcm, bool flag ){
    Lcm = Lcm*num[ id ] / gcd( Lcm,num[ id ] );
    if( flag ) ans += n/Lcm;
    else ans -= n/Lcm;
    for( int i=id+1;i<= N;i++ ){
        dfs( i,Lcm,!flag );
    }
    return;
}

int main() {
    ios::sync_with_stdio(false);
    cin >> N;
    for (int i = 1; i <= N; i++) cin >> num[i];
    int D;
    cin >> D;
    for (int i = 1; i <= D; i++) {
        LL l, r;
        cin >> l >> r;
        LL ansr = 0, ansl = 0;
        ans = 0;
        n = r;
        for (int i = 1; i <= N; i++) dfs(i, num[i], true);
        ansr = ans;
        ans = 0;
        n = l - 1;
        for (int i = 1; i <= N; i++) dfs(i, num[i], true);
        ansl = ans;
        cout << ansr - ansl << endl;
    }
    return 0;
}

E题

从难度来看这题算是简单的规律题。。可是比赛中却没敢去写。

通过数列的前几项可以很自然的和fib数联系在一起。我们需要做的就是统计每个二进制位上的值。

通过本题学习了bitset的使用，很好用，支持两个bitset间进行位运算，很方便。

#include <cmath>
#include <bitset>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

const int MOD = 1e9 + 7;
const int MAX_N = 500050;
typedef long long LL;
#define rep(i, n) for (int i = (1); i <= (n); i++)
const int NUM = 100;
LL fib[NUM], sum[NUM];

LL mod_pow(LL a, LL b) {
    LL res = 1;
    while (b) {
        if (b & 1) res = (res * a) % MOD;
        a = (a * a) % MOD;
        b >>= 1;
    }
    return res;
}

bitset<NUM> solve(LL x) {
    bitset<NUM> res;
    while (x) {
        int pos = upper_bound(sum, sum + NUM, x) - sum - 1;
        x -= sum[pos];
        res.set(pos);
    }
    return res;
}

int main() {
    fib[0] = 0; fib[1] = 1;
    sum[0] = 1; sum[1] = 2;  
    for (int i = 2; i < NUM; i++) {
        fib[i] = fib[i - 2] + fib[i - 1];
        sum[i] = sum[i - 1] + fib[i];
    }
    
    ios_base::sync_with_stdio(false);
    int N;
    cin >> N;
    bitset<NUM> res;
    rep (i, N) {
        LL x;
        cin >> x;
        res ^= solve(x);
    }
    LL ans = 0;
    for (int i = 0; i < res.size(); i++) if (res.test(i)) ans = (ans + mod_pow(2, i)) % MOD;
    cout << ans << endl;
    return 0;
}

F题

待续。。

G题

待续。。

H题

待续。。