ASC 46

最新推荐文章于 2021-02-20 20:41:03 发布

原创最新推荐文章于 2021-02-20 20:41:03 发布 · 499 阅读

0 ·

CC 4.0 BY-SA版权

ASC 专栏收录该内容

6 篇文章

订阅专栏

A Astronomy Problem

直接双指针。

B Bipartite Bicolored Graphs

#include<cstdio>
#define LL long long
#define MAXN 301
const int MOD = 175781251;
int c[MAXN][MAXN], ans[MAXN][MAXN], inv[MAXN];
int pow2[MAXN], pow3[MAXN*MAXN];
inline int mul(int x, int y){ return (LL)x*y%MOD; }
inline int add(int x, int y){ return (x + y) % MOD; }
inline int sub(int x, int y){ return (x + MOD - y) % MOD; }
void init()
{
    for (int i = 0; i < MAXN; i++){
        c[i][0] = c[i][i] = 1;
        for (int j = 1; j < i; j++)
            c[i][j] = add(c[i - 1][j - 1], c[i - 1][j]);
    }
    pow2[0] = pow3[0] = 1;
    for (int i = 1; i < MAXN; i++)
        pow2[i] = mul(pow2[i - 1], 2);
    for (int i = 1; i < MAXN*MAXN; i++)
        pow3[i] = mul(pow3[i - 1], 3);
    inv[1] = 1;
    for (int i = 2; i < MAXN; i++)
        inv[i] = mul(inv[MOD%i], MOD - MOD / i);
    for (int n = 1; n < MAXN; n++){
        for (int i = 0; i <= n; i++)
            ans[1][n] = add(ans[1][n], mul(c[n][i], pow3[i*(n - i)]));
        ans[1][n] = mul(ans[1][n], inv[2]);
        for (int i = 2; i <= n; i++){
            for (int k = n - i + 1; k; k--)
                ans[i][n] = add(ans[i][n], mul(mul(c[n][k], ans[1][k]), ans[i - 1][n - k]));
            ans[i][n] = mul(ans[i][n], inv[i]);
            ans[1][n] = sub(ans[1][n], mul(pow2[i - 1], ans[i][n]));
        }
    }
}
int main()
{
    freopen("bipartite.in", "r", stdin);
    freopen("bipartite.out", "w", stdout);
    int n;
    init();
    while (scanf("%d", &n) == 1 && n){
        int sum = 0;
        for (int i = 1; i <= n; i++)
            sum = add(sum, ans[i][n]);
        printf("%d\n", sum);
    }
}

E Ebola Virus

#include<bits/stdc++.h>

using namespace std;

const double eps=1e-10;
const double pi=3.1415926535897932384626433832795;
const double eln=2.718281828459045235360287471352;

#define LL long long
#define IN freopen("ebola.in", "r", stdin)
#define OUT freopen("ebola.out", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr(x) printf("Case %d: ",x)
#define prn(x) printf("Case %d:\n",x)
#define prr(x) printf("Case #%d: ",x)
#define prrn(x) printf("Case #%d:\n",x)
#define lowbit(x) (x&(-x))
#define fi first
#define se second
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

const int maxn=3005;
LL g[maxn][maxn];
LL a[maxn],f[maxn];
int n;

int main()
{
    IN;OUT;
    while(scanf("%d",&n)==1 && n)
    {
        a[n+1]=0;
        for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
        for(int i=n;i>=1;i--)a[i]+=a[i+1];
        for(int i=1;i<=n;i++)
        {
            f[i]=1e18;
            for(int j=1;j<=n;j++)g[i][j]=1e18;
        }
        for(int i=1;i<=n;i++)g[i][i]=a[i+1];
        for(int len=2;len<=n;len++)
            for(int i=1;i+len-1<=n;i++)
            {
                int j=i+len-1;
                //min(g[i+1][j]+2ll*a[i+1]+a[j+1],g[i+1][j]+2ll*a[j+1]+a[i+1]+(a[i]-a[i+1]));
                LL x=g[i+1][j]+a[i+1]+a[j+1]+a[i+1];
                LL y=g[i+1][j]+a[i+1]+2ll*a[j+1]+1ll*(3*(j-i))*(a[i]-a[i+1]);
                //if(i==1 && j==2)printf("!!! %lld %lld\n",x,y);
                g[i][j]=min(x,y);
            }
        f[0]=0;
        f[1]=a[2];
        for(int i=2;i<=n;i++)
        {
            f[i]=f[i-1]+a[i]+a[i+1];
            f[i]=min(f[i],g[1][i]+1ll*a[i+1]*i);
            f[i]=min(f[i],g[1][i]+1ll*a[i+1]*(i-1));
            //printf("%d %lld\n",i,f[i]);
            for(int k=1;k<i;k++)
                f[i]=min(f[i],f[k]+a[k+1]+g[k+1][i]+1ll*a[i+1]*(i-k-1));
        }
        LL ans=f[n];
        for(int i=1;i<n;i++)
            ans=min(ans,f[i]+g[i+1][n]+a[i+1]);    
        printf("%lld\n",ans);
    }
    return 0;
}

F Figure Skating

矩阵快速幂

#include<cstdio>
#include<cstring>
#define LL long long
#define MAXN 5
const int MOD=998244353;
struct Matrix{
    int a[MAXN][MAXN], n;
    Matrix(int n = 0) :n(n){}
    void zero(){
        for (int i = 0; i < n; i++)
            memset(a[i], 0, sizeof(int)*n);
    }
    void one(){
        zero();
        for (int i = 0; i < n; i++)a[i][i] = 1;
    }
    Matrix operator * (const Matrix& x)const{
        Matrix r(n); r.zero();
        for (int i = 0; i < n; i++){
            for (int j = 0; j < n; j++){
                if (a[i][j]){
                    for (int k = 0; k < n; k++)
                        r.a[i][k] = (r.a[i][k] + (LL)a[i][j] * x.a[j][k]) % MOD;
                }
            }
        }
        return r;
    }
    Matrix operator ^ (LL k)const{
        Matrix r(n), t = *this; r.one();
        for (; k; k >>= 1){
            if (k & 1)r = r * t;
            t = t * t;
        }
        return r;
    }
};
int main()
{
    freopen("figure.in","r",stdin);
    freopen("figure.out","w",stdout);
    int n;
    Matrix m(3);
    m.a[0][0]=1;m.a[0][1]=1;m.a[0][2]=0;
    m.a[1][0]=1;m.a[1][1]=2;m.a[1][2]=1;
    m.a[2][0]=0;m.a[2][1]=1;m.a[2][2]=2;
    while(scanf("%d",&n)==1&&n){
        Matrix t=m^n;
        printf("%d\n",t.a[0][0]);
    }
}

G Game of Col on Bamboo Forests

#include<bits/stdc++.h>

using namespace std;

const double eps=1e-10;
const double pi=3.1415926535897932384626433832795;
const double eln=2.718281828459045235360287471352;

#define LL long long
#define IN freopen("game.in", "r", stdin)
#define OUT freopen("game.out", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr(x) printf("Case %d: ",x)
#define prn(x) printf("Case %d:\n",x)
#define prr(x) printf("Case #%d: ",x)
#define prrn(x) printf("Case #%d:\n",x)
#define lowbit(x) (x&(-x))
#define fi first
#define se second
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

const int mod=242121643;
int c[105][105];
int n,m,s0,s1;

int main()
{
    IN;OUT;
    c[0][0]=c[1][1]=c[1][0]=1;
    for(int i=2;i<=100;i++)
    {
        c[i][0]=c[i][i]=1;
        for(int j=1;j<i;j++)
            c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
    }
    while(scanf("%d%d",&n,&m)==2 && (n+m))
    {
        s0=s1=0;
        for(int i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            if(x==1)s1++;else s0++;
        }
        int ans=0;
        for(int i=1;i<=min(s1,m);i+=2)
        {
            int now=1ll*c[s1][i]*c[s0][m-i]%mod;
            ans=(ans+now)%mod;
        }
        printf("%d\n",ans);
    }
    return 0;
}

J Jingles of a String

对每个i求以i为右节点的j个字母的最长串

#include<bits/stdc++.h>

using namespace std;

const double eps=1e-10;
const double pi=3.1415926535897932384626433832795;
const double eln=2.718281828459045235360287471352;

#define LL long long
#define IN freopen("jingles.in", "r", stdin)
#define OUT freopen("jingles.out", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr(x) printf("Case %d: ",x)
#define prn(x) printf("Case %d:\n",x)
#define prr(x) printf("Case #%d: ",x)
#define prrn(x) printf("Case #%d:\n",x)
#define lowbit(x) (x&(-x))
#define fi first
#define se second
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

int T,mod=4000007;
char s[100005];
int h[4000008];
int a[27],v[27];
LL ans[4000008];
int fl[4000008],nu[4000008],pos[27];
int n;
pii c[27];
int fk[4000008],cc;

int has(int x)
{
    int now=x%mod;
    while(h[now]!=x && fl[now]==T+1)now=(now+1)%mod;
    h[now]=x;
    if(fl[now]!=T+1)
    {
        nu[now]=0;ans[now]=0;
        fk[++cc]=now;
        fl[now]=T+1;
    }
    return now;
}

int main()
{
    IN;OUT;
    scanf("%d",&T);
    c[0]=mp(0,0);
    while(T--)
    {
        scanf("%s",s+1);
        n=strlen(s+1);
        cc=0;
        for(int i=1;i<=26;i++)v[i]=rand()%1000+1,pos[i]=0;
        for(int i=1;i<=n;i++)
        {
            int x=s[i]-96;
            pos[x]=i;
            for(int i=1;i<=26;i++)c[i]=mp(pos[i],i);
            sort(c+1,c+27);
            int now=0;
            for(int j=26;j>=1;j--)
            {
                if(c[j].fi<=0)break;
                now|=1<<(c[j].se-1);
                int u=has(now);
                nu[u]=max(nu[u],i-c[j-1].fi);
                ans[u]=1ll*nu[u]*(26-j+1);
            }
        }
        LL sum=0;
        for(int i=1;i<=cc;i++)sum+=ans[fk[i]];
        printf("%d %lld\n",cc,sum);
    }
    return 0;
}