小鸡腿U T9 Dhaka2015

Sum of MSLCM

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> #include<vector>#include<string>#include<queue>#include<stack>#include<map>#include<sstream>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} int main(){//  freopen("E.in","r",stdin);//  freopen(".out","w",stdout);    ll n;     while(cin>>n&&n) {        ll ans=0;        for(ll i=1;i<=n;) {            ll t=n/(n/i);            ll p=(i+t)*(t-i+1)/2;            ans+=n/i*p;            i=t+1;        }        ans--;        cout<<ans<<endl;    }    return 0;}

Numbered Cards

You have N cards and each has an unique number between 1 and N written on it. In how many
ways can you select a non-empty subset of the cards such that the number written on any two of your
selected cards don’t have any common digits?
For example, when N = 12, {1, 2, 3}, {2, 11}, {3, 4, 5, 6, 7, 8, 9, 12} are some valid selections. But
{1, 2, 10}, {2, 5, 12} are not allowed.

数位dp

The Hypnotic Spirals

阿基米德螺旋线到θ角为止所围面积=a2θ32
我特判了：
1条直线的情况
有一条ray为x=0的情况
注意一开始的一圈
注意跨越x=0轴部分的面积要特判
注意n=0特判

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> #include<vector>#include<string>#include<queue>#include<stack>#include<map>#include<sstream>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (20+10)#define MAXM (1000+10)ld PI = 3.141592653589793238462643383;double a,xita[MAXN];int n,m;pair<double,double> q[MAXM];double calcS(double xit) {    return xit*xit*xit;} double calc() {    ld k=2*PI*a;    vector<pi > v;    bool b=0;    Rep(i,m) {        double xit=q[i].se,r=q[i].fi;        if (0<xit && xit < xita[0] && r<a*xit ) {            b=1; continue;        }        int bl;        if (xit<xita[0] || xit>xita[n-1]) bl=n-1;        else bl=lower_bound(xita,xita+n,q[i].se)-xita-1;        int p;        if (r<a*xit) p=0;        else p=(int)(0.5+ceil((r-a*xit)/k));        if (bl==n-1) {            if (0<=xit && xit<xita[0] ) p--;        }        v.pb(mp(bl,p));//      cout<<bl<<' '<<p<<endl;    }    sort(v.begin(),v.end());    int sz=unique(v.begin(),v.end())-v.begin();    double ans=0;    if (b) ans+=xita[0]*xita[0]*xita[0];    Rep(i,sz) {        int lx = v[i].fi,rx=(v[i].fi+1)%n;              double l = xita[lx], r = xita[rx];        if (lx==n-1) r+=2*PI;        double p=v[i].se;        {            double l1 = l + (p-1)*2*PI , r1 = r+ (p-1)*2*PI;            double l2 = l1 + 2*PI , r2 = r1 + 2*PI;            l1=max(l1,0.); l2=max(l2,0.);            r1=max(r1,0.); r2=max(r2,0.);            ans+=r2*r2*r2-l2*l2*l2;            ans-=r1*r1*r1-l1*l1*l1;        }    }    ans=ans*a*a/6;    return ans;}double calc2() {    vector< int > v;    double k=2*PI*a;    Rep(i,m) {        double xit=q[i].se,r=q[i].fi;        if (r<a*xit &&0<xit && xit<xita[0]) v.pb(0);        else {            int p;            {                p=(0.5+ceil((r-a*xit)/k));                if (0<=xit && xit<xita[0] ) p--;                v.pb(p+1);            }        }    }       sort(v.begin(),v.end());    int sz=unique(v.begin(),v.end())-v.begin();    double ans=0;       Rep(i,sz) {        int c=v[i];        if (v[i]==0) {            ans+=calcS(xita[0]);        } else {            double r=xita[0]+2*PI*c,l1=r-2*PI,l2=l1-2*PI;            r=max(r,0.); l1=max(l1,0.); l2=max(l2,0.);            ans+=r*r*r-l1*l1*l1;            ans-=l1*l1*l1-l2*l2*l2;                 }    }    return ans*a*a/6;}int main(){//  freopen("J.in","r",stdin);//  freopen(".out","w",stdout);    int T=read();    For(kcase,T) {        scanf("%lf",&a);        n=read();        Rep(i,n) scanf("%lf",&xita[i]);        sort(xita,xita+n);        m=read();        Rep(i,m) {            scanf("%lf%lf",&q[i].fi,&q[i].se);        }        printf("Spiral %d:",kcase);        if (n==0) {            printf(" -1\n");        } else if (n==1) {            printf(" %.4lf liters\n",calc2()/10);        } else {            printf(" %.4lf liters\n",calc()/10);        }    }    return 0;}