牛客小白月赛13 解题报告

暂时咕着，慢慢补...赛时几题先放上来

A

打个表即可

#include<bits/stdc++.h>
using namespace std;
int main() {
    long long n;
    scanf("%lld",&n);
    if(n & 1) puts("1"); else puts("-1");
}

B

枚举端点，hash判断即可。

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
#define N 10100
#define base 13131
 
char s[N];
ull h1[N], p[N], h2[N];
int n, ans = 0;
 
ull gh1(int l, int r) { return h1[r] - h1[l - 1] * p[r - l + 1]; }
ull gh2(int l, int r) { return h2[l] - h2[r + 1] * p[r - l + 1]; }
 
int main() {
    scanf("%s", s + 1); p[0] = 1;
    n = strlen(s + 1);
    for(int i = 1; i <= n; ++i) s[i + n] = s[i];
    n *= 2;
    for(int i = 1; i <= n; ++i) {
        h1[i] = h1[i - 1] * base + s[i];
        p[i] = p[i - 1] * base;
    }
    for(int i = n; i; i--) h2[i] = h2[i + 1] * base + s[i];
    for(int i = 1; i <= n; ++i) {
        for(int j = i; j - i + 1 <= n / 2 && j <= n; ++j) {
            int mid = (i + j) >> 1;
            if((j - i + 1) & 1 && gh1(i, mid - 1) == gh2(mid + 1, j)) ans = max(ans, j - i + 1);
            if(!((j - i + 1) & 1) && gh1(i, mid) == gh2(mid + 1, j)) ans = max(ans, j - i + 1);
        }
    }
    printf("%d
", ans);
}

C

考虑答案长什么样子

[frac{钱数大于3*n的方案数}{总方案数} ]

这个方案数dp一下就可以了。类似背包。
然后总方案数是什么呢？考虑每次一共有4种选择，相乘即(4^n)。
用个gcd约分即可。

#include <bits/stdc++.h>
using namespace std;
#define ll long long

ll f[40][130];
int n;

int main() {
	cin >> n;
	f[0][0] = 1;
	for(int i = 1; i <= n; ++i) {
		for(int j = 1; j <= 4 * n; ++j) 
			for(int v = 1; v <= 4; ++v) 
				if(j - v >= 0) f[i][j] += f[i - 1][j - v];
	}
	ll m1 = 0;
	for(int i = 3 * n; i <= 4 * n; ++i) m1 += f[n][i];
	ll m2 = pow(4, n), g = __gcd(m1, m2);
	printf("%lld/%lld
", m1 / g, m2 / g);
}

D

维护前后缀和，枚举不选点

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 5000010
 
ll a[N], l[N], r[N];
 
int main() {
    int n; cin >> n;
    for(int i = 1; i <= n; ++i) scanf("%lld", &a[i]), l[i] = l[i - 1] | a[i];
    for(int i = n; i; --i) r[i] = r[i + 1] | a[i];
    ll ans = 0;
    for(int i = 1; i <= n; ++i) {
        ans = max(ans, l[i - 1] | r[i + 1]);
    }  
    printf("%lld
", ans);
}

E

倍增floyd，转移的时候相乘即可。注意细节。

#include <cstdio>
#include <algorithm>
#include <cstring>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('
')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 110
const ll mod = 1000000007;

ll K,m,s,n;

struct mat {
    int m[N][N];
    mat operator * (const mat x) const {
        mat c; memset(c.m, 0, sizeof(c.m));
        for(int k = 1; k <= n; ++k) {
            for(int i = 1; i <= n; ++i) {
                for(int j = 1; j <= n; ++j) {
                    c.m[i][j] += m[i][k] * x.m[k][j] % mod;
                    c.m[i][j] %= mod;
                }
            }
        }
        return c;
    }
}d[32];

int main() {
    in(n);in(m);in(K);in(s); int tot = 0;
    for(int i = 1; i <= m; ++i) {
    	int x, y; in(x), in(y);
    	d[0].m[x][y]++;
    }
    for(int i = 1; (1 << i) <= K; ++i) d[i] = d[i - 1] * d[i - 1];
    mat ans;
    --K; ans = d[0];
    for(int i = 0; (1 << i) <= K; ++i) if((K>>i)&1) ans = ans * d[i];
    ll sum = 0;
	for(int i = 1; i <= n; ++i) {
		if(i == s) continue;
		if(ans.m[s][i]!=-1) sum += ans.m[s][i] % mod;
		sum %= mod;
	}
	outn(sum);
}

H

单调栈板子。

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define N 1000100
int a[N], n, k, h[N], w[N], s[N];
ll ans = 0;
 
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for(int i = 1; i <= n; ++i) scanf("%d", &h[i]), ans = max(ans, (ll)h[i] * a[i]);
    int top = 0;
    for(int i = 1; i <= n; ++i) {
        if(a[i] > h[s[top]]) {
            s[++top] = i, w[top] = 1;
        } else {
            ll W = 0;
            while(top && h[s[top]] > h[i]) {
                W += w[top];
                ans = max(ans, (ll)W * h[s[top]]);
                --top;
            }
            s[++top] = i; w[top] = W + a[i];
        }
    }
    printf("%lld
", ans);
}

I

做一下nim，如果必败就枚举要取哪个，如果都不行那就还是必败

#include <bits/stdc++.h>
using namespace std;
#define N 100010
int a[N], s, n, k;
  
int main() {
    cin >> n >> k;
    for(int i = 1; i <= n; ++i) {
        cin >> a[i]; s ^= a[i];
    }
    if(s) return puts("YES"), 0;
    for(int i = 1; i <= n; ++i) {
        int now = s ^ a[i];
        if(a[i] >= k && now ^ (a[i] - k)) return puts("YES"), 0;
    }
    puts("NO");
}

J

套路莫反。这里放上推导。复杂度(O(sqrt{n}))

[egin{aligned} &sum sum (i,j)^2\ &=sum^n d^2sum sum[(i,j)=d]\ &=sum^n d^2 sum_{i=1}^{n/d}sum_{j=1}^{m/d}[(i,j)=1]\ &=sum^n d^2 sum_{k=1}^{n/d}mu(k)frac{n}{kd}frac{m}{kd}\ &T = kd\ &=sum_{T=1}^nfrac{n}{T}frac{m}{T}sum_{k|T}mu(k)(frac{T}{k})^2\ &id^2*mu\ &=F(1)=1, F(p)=p^2-1, F(p^2)=F(p)*p^2 end{aligned} ]

[egin{aligned} &sum_{k|T}mu(k)(frac{T}{k})^2\ &mu*id^2\ &F(1)=1\ &F(p)=1*(p/1)^2+(-1)*(p/p)^2=p^2-1\ &F(p^2)=1*(p^2/1)^2+(-1)*p^2+0=(p^2)^2-p^2=(p^2-1)*p^2 end{aligned} ]

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define N 1000100
#define M 1000000
const ll mod = 1000000007;
 
ll n, m;
int p[N], cnt, vis[N];
ll F[N];
 
void init() {
    F[1] = 1;
    for(int i = 2; i <= M; ++i) {
        if(!vis[i]) p[++cnt] = i, F[i] = ((ll)i * i - 1ll) % mod;
        for(int j = 1; i * p[j] <= M && j <= cnt; ++j) {
            vis[i * p[j]] = 1;
            if(i % p[j] == 0) {
                F[i * p[j]] = F[i] * (p[j] * p[j] % mod) % mod;
                break;
            }
            F[i * p[j]] = (ll)F[i] * F[p[j]] % mod;
        }
    }
    for(int i = 1; i <= M; ++i) F[i] += F[i - 1], F[i] %= mod;
}
 
int main() {
    init();
    scanf("%lld%lld", &n, &m);
    if(n > m) swap(n, m);
    ll ans = 0;
    for(ll l = 1, r; l <= n; l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ans = (ans + (ll)(n / l) * (m / l) % mod * ((F[r] - F[l - 1] + mod) % mod) % mod) % mod;
    }
    printf("%lld
", (ans % mod + mod) % mod);
}