ASC 46

A Astronomy Problem

直接双指针。

B Bipartite Bicolored Graphs

#include<cstdio>#define LL long long#define MAXN 301const 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 secondtypedef 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 5const 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 secondtypedef 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 secondtypedef 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;}