【NOIP2017】Day7

Solution

T1：
推了一个考试的数学公式，贼复杂，事实上正解也挺复杂的，什么二次函数求和？此处%ljn大佬的思路+精妙的代码，思路：6个点最多分成5的区域，注意左闭右开后枚举每个机器人在那个区间，在分类讨论1.3个在同一区间2.两个在同一区间3.在不同区间，分别计算，这个好算多了
T2：
数位DP，考试前讲过，A了，CF上的题
http://www.cnblogs.com/vongang/p/3141273.html
http://www.cnblogs.com/hxer/p/5169877.html
http://blog.csdn.net/keshuai19940722/article/details/19605745
http://blog.csdn.net/qq_33184171/article/details/52332586
贴几份吧，不想讲了
T3：
其实是水题，第一问可以直接算出各个字符数在计算，
40%：预处理字符串中各个字符的个数，计算O（N^2*62^2）
100%：预处理字符串与各个字符之间的贡献，计算O（N^2*62）

CODE

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>using namespace std;#define ll long longint l[4],r[4],a[10],cnt;double C2(double x){    return x*(x-1)/2.0;}double C3(double x){    return x*(x-1)*(x-2)/6.0;}int main(){    freopen("jump.in","r",stdin);    freopen("jump.out","w",stdout);    int T;    scanf("%d",&T);    while (T--)    {        cnt=0;        double ans=0;        for (int i=1;i<=3;i++)        {            scanf("%d%d",&l[i],&r[i]);            a[++cnt]=l[i]; a[++cnt]=r[i]+1;        }        sort(a+1,a+1+cnt);        cnt=unique(a+1,a+1+cnt)-a-1;        for (int i=1;i<=cnt-1;i++)            for (int j=1;j<=i;j++)                for (int k=1;k<=j;k++)                    if (l[1]<=a[i] && a[i+1]-1<=r[1] && l[2]<=a[j] && a[j+1]-1<=r[2] && l[3]<=a[k] && a[k+1]-1<=r[3])                    {                        double n=a[i+1]-a[i],m=a[j+1]-a[j],p=a[k+1]-a[k];                        if (i==j && j==k)                            ans+=C3(n)+C2(n)*2.0+(double)n;                        else if (i!=j && j!=k && i!=k)                            ans+=n*m*p;                        else if (i==j)                            ans+=C2(n)*p+n*p;                        else if (j==k)                            ans+=C2(m)*n+n*m;                    }        for (int i=1;i<=3;i++) ans/=(double)(r[i]-l[i]+1);        printf("%.9lf\n",ans);              }       return 0;}

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>using namespace std;#define ll long longll pow10[20],Pow[20],dp[20][2][2520][50];int num[20],flcm[512],Map[2521],Map2[50],tot;int gcd(int a,int b){    return b==0?a:gcd(b,a%b);}int lcm(int a,int b){    if (a==0) return b;    else if (b==0) return a;    else return a/gcd(a,b)*b;}ll solve(ll x){    int len;    ll ans=0;    if (x==1e18) ans=1,x--;    for (len=0;len<=18 && pow10[len]<=x;len++);    for (int i=len;i>0;i--) num[i]=x%pow10[len-i+1]/pow10[len-i];    memset(dp,0,sizeof(dp));    dp[0][1][0][0]=1;    for (int i=0;i<=len;i++)        for (int j=0;j<2;j++)            for (int k=0;k<2520;k++)                for (int l=0;l<tot;l++)                {                    ll p=dp[i][j][k][l];                    int pp=Map2[l];                    if (!p) continue;                    if (i==len)                     {                        if (k % pp==0) ans+=p;                        continue;                    }                    if (j==0)                    {                        for (int q=0;q<=9;q++)                            dp[i+1][0][(k+q*Pow[len-i-1])%2520][Map[lcm(pp,q)]]+=p;                                         }                    else                    {                        for (int q=0;q<num[i+1];q++)                            dp[i+1][0][(k+q*Pow[len-i-1])%2520][Map[lcm(pp,q)]]+=p;                        dp[i+1][1][(k+num[i+1]*Pow[len-i-1])%2520][Map[lcm(pp,num[i+1])]]+=p;                    }                               }    return ans;}void prepare(){    pow10[0]=Pow[0]=1;    for (int i=1;i<=18;i++) pow10[i]=pow10[i-1]*10,Pow[i]=pow10[i]%2520;    for (int i=1;i<(1<<9);i++)        for (int j=1;j<=9;j++)            if (i & (1<<j-1))                flcm[i]=lcm(flcm[i-(1<<j-1)],j);    for (int i=1;i<(1<<9);i++)        if (!Map[flcm[i]]) Map[flcm[i]]=tot++,Map2[tot-1]=flcm[i];}int main(){    freopen("number.in","r",stdin);    freopen("number.out","w",stdout);    prepare();    int T;    scanf("%d",&T);    while (T--)    {        ll a,b;        scanf("%lld%lld",&a,&b);        printf("%lld\n",solve(b)-solve(a-1));    }       return 0;}

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>using namespace std;#define ll long long#define MOD 1000000007char str1[10],str2[10],str[1000010];ll a[100][100],sum[1010][100],cnt[1010][100];int n,m;int id(char ch){    if (ch>='0' && ch<='9') return ch-'0'+1;    else if (ch>='A' && ch<='Z') return ch-'A'+11;    else if (ch>='a' && ch<='z') return ch-'a'+37;}void solve(){    ll ans=0,ans2=0;    for (int i=1;i<=n;i++)        for (int j=i;j<=n;j++)        {            ll tot=0;            for (int k=1;k<=62;k++)                tot=(tot+sum[i][k]*cnt[j][k])%MOD;            ans=(ans+tot)%MOD;            ans2=(ans2+tot*tot)%MOD;        }    cout<<ans<<endl<<ans2<<endl;    }void ReadInfo(){    scanf("%d%d",&n,&m);    for (int i=1;i<=m;i++)    {        int x;        scanf("%s",str1);        scanf("%s",str2);        scanf("%d",&x);        int y=id(str1[0]),z=id(str2[0]);        a[y][z]=a[z][y]=x;    }    for (int i=1;i<=n;i++)    {        scanf("%s",str+1);          int len=strlen(str+1);        for (int j=1;j<=len;j++)            cnt[i][id(str[j])]++;        for (int j=1;j<=62;j++)            for (int k=1;k<=62;k++)                sum[i][j]=(sum[i][j]+cnt[i][k]*a[j][k])%MOD;    }}int main(){    freopen("string.in","r",stdin);    freopen("string.out","w",stdout);    ReadInfo();    solve();    return 0;}