Codeforces Round #258 (Div. 2)-(A,B,C,D,E)

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http://blog.csdn.net/rowanhaoa/article/details/38116713

A:Game With Sticks

水題。。。每次操作，都會拿走一個橫行，一個豎行。

所以一共會操作min（橫行，豎行）次。

#include<stdio.h>#include<iostream>#include<stdlib.h>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<map>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define mem(a,b) (memset(a),b,sizeof(a))#define lmin 1#define rmax n#define lson l,(l+r)/2,rt<<1#define rson (l+r)/2+1,r,rt<<1|1#define root lmin,rmax,1#define now l,r,rt#define int_now int l,int r,int rt#define INF 99999999#define LL __int64#define mod 1000000009#define eps 1e-6#define zero(x) (fabs(x)<eps?0:x)#define maxn 330000int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        int y=min(n,m);        if(y%2)cout<<"Akshat"<<endl;        else cout<<"Malvika"<<endl;    }    return 0;}

B：Sort the Array

給數組中的每一個數字標號，標記他應該出現在哪一個位置。

然後從頭往後找，如果當前位置的數字不是應該在這個位置的數字。

那麼就找到應該在這個位置的數位位置，然後翻轉。如果還不是按順序排的話，就輸出no，否則輸出yes。

#include<stdio.h>#include<iostream>#include<stdlib.h>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<map>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define mem(a,b) (memset(a),b,sizeof(a))#define lmin 1#define rmax n#define lson l,(l+r)/2,rt<<1#define rson (l+r)/2+1,r,rt<<1|1#define root lmin,rmax,1#define now l,r,rt#define int_now int l,int r,int rt#define INF 99999999#define LL __int64#define mod 1000000009#define eps 1e-6#define zero(x) (fabs(x)<eps?0:x)#define maxn 110000struct list{    int x;    int id;    int y;}node[maxn];int cmp1(list a,list b){    return a.x<b.x;}int cmp2(list a,list b){    return a.id<b.id;}int main(){    int n,m;    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)        {            scanf("%d",&node[i].x);            node[i].id=i;        }        sort(node+1,node+n+1,cmp1);        for(int i=1;i<=n;i++)node[i].y=i;        sort(node+1,node+n+1,cmp2);        int st,ed,i;        st=ed=-1;        for(i=1;i<=n;i++)        {            if(node[i].y!=i)            {                if(st==-1)                {                    int j,k;                    for(j=i+1;j<=n;j++)                    {                        if(node[j].y==i)break;                    }                    for(k=j;k>=i;k--)                    {                        if(node[k].y!=i+(j-k))break;                    }                    if(k>=i)break;                    st=i;                    ed=j;                    i=j+1;                }                else break;            }        }        if(i<=n)cout<<"no"<<endl;        else        {            cout<<"yes"<<endl;            if(st!=-1)cout<<st<<" "<<ed<<endl;            else cout<<"1 1"<<endl;        }    }    return 0;}

C：Predict Outcome of the Game

枚舉兩個差值的加號或減號。

對於每一種情況：

可以算出A，B,C最少贏幾個球。然後看看當前的贏球數是不是符合K。

然後看一下差值是否可以用(n-k)消除掉。

#include<stdio.h>#include<iostream>#include<stdlib.h>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<map>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define mem(a,b) (memset(a),b,sizeof(a))#define lmin 1#define rmax n#define lson l,(l+r)/2,rt<<1#define rson (l+r)/2+1,r,rt<<1|1#define root lmin,rmax,1#define now l,r,rt#define int_now int l,int r,int rt#define INF 99999999#define LL __int64#define mod 1000000009#define eps 1e-6#define zero(x) (fabs(x)<eps?0:x)#define maxn 110000struct list{    int x;    int id;    int y;}node[maxn];int cmp1(list a,list b){    return a.x<b.x;}int cmp2(list a,list b){    return a.id<b.id;}int pan(LL n,LL k,LL d1,LL d2){    LL a,b,c;    a=0;    b=a+d1;    c=b+d2;    LL d;    d=min(a,min(b,c));    if(d<0)    {        d=-d;        a+=d;        b+=d;        c+=d;    }    d=max(a,max(b,c));    LL cha=0;    cha=a+b+c;    cha=k-cha;    if(cha<0)return 0;    if(cha%3)return 0;    cha=0;    cha+=d-a;    cha+=d-b;    cha+=d-c;    LL cun=n-k;    cun=cun-cha;    if(cun<0)return 0;    if(cun%3==0)    {     //   cout<<n<<" "<<k<<" "<<d1<<" "<<d2<<endl;        return 1;    }    else return 0;}void dos(){    LL n,k,d1,d2;    cin>>n>>k>>d1>>d2;    LL a,b,c;    a=0;    if(pan(n,k,d1,d2))    {        cout<<"yes"<<endl;        return;    }    if(pan(n,k,-d1,d2))    {        cout<<"yes"<<endl;        return;    }    if(pan(n,k,d1,-d2))    {        cout<<"yes"<<endl;        return;    }    if(pan(n,k,-d1,-d2))    {        cout<<"yes"<<endl;        return;    }    cout<<"no"<<endl;}int main(){    int t;    LL n,k,d1,d2;    while(~scanf("%d",&t))    {        while(t--)        {            dos();        }    }    return 0;}

D：Count Good Substrings

假如字串為：abbabbbaaa

我們把這個字串如下記錄: 

     字串： a b a b a

數組num ：1 2 1 3  3

數組的每一項代表這個字串這個位置的字元是由幾個字元壓縮成的。

對於迴文串的兩邊都是a的情況：

預先處理從起點走奇數步可到達多少a，走偶數步，可到達多少a。

然後從第一個a往後走，可在O（1）的複雜度內得出當前a的為迴文串的左邊，一共有幾個奇數子串，幾個偶數子串。

對於迴文串的兩邊都是b的情況，類似與兩邊都是a的情況。

#include<stdio.h>#include<iostream>#include<stdlib.h>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<map>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define mem(a,b) (memset(a),b,sizeof(a))#define lmin 1#define rmax n#define lson l,(l+r)/2,rt<<1#define rson (l+r)/2+1,r,rt<<1|1#define root lmin,rmax,1#define now l,r,rt#define int_now int l,int r,int rt#define INF 99999999#define LL __int64#define mod 1000000009#define eps 1e-6#define zero(x) (fabs(x)<eps?0:x)#define maxn 110000LL a[maxn];char str[maxn];vector<char>vec;int num[maxn];LL s1,s2;LL cal(int x,int y){    if(x<0)return 0;    LL a1,an,n;    if(y==1)    {        if(x<2)return 0;        an=x-1;        if(x%2)a1=2;        else a1=1;        n=(an-a1)/2+1;        return n*(a1+an)/(LL)2;    }    else    {        an=x;        if(x%2)a1=1;        else a1=2;        n=(an-a1)/2+1;        return n*(a1+an)/(LL)2;    }}void dos(int x){    int n=vec.size();    LL even,odd;    even=odd=0;    LL sum=0;    for(int i=1; i<=n; i++)    {        if(i%2==x)        {            LL a,b;            a=num[i]/2;            b=num[i]-a;            if(sum%2)swap(a,b);            even+=a;//偶            odd+=b;//奇        }        sum+=num[i];    }    sum=0;    for(int i=x;i<=n;i+=2)    {        LL a,b;        a=num[i]/2;        b=num[i]-a;        if(sum%2)swap(even,odd);        sum=0;        even-=a;        odd-=b;        sum+=num[i];        if(sum%2)swap(even,odd);        sum=0;        s1+=b*odd;        s1+=a*even;        s2+=b*even;        s2+=a*odd;        s1+=cal(num[i],1);        s2+=cal(num[i],2);        sum+=num[i+1];    }}int main(){    int n;    LL s;    while(~scanf("%s",str))    {        vec.clear();        int len=strlen(str);        int s=1;        for(int i=1; i<len; i++)        {            if(str[i]!=str[i-1])            {                vec.push_back(str[i-1]);                num[vec.size()]=s;                s=1;            }            else s++;        }        vec.push_back(str[len-1]);        num[vec.size()]=s;        s1=s2=0;        dos(1);        dos(0);        cout<<s1<<" "<<s2<<endl;    }    return 0;}

E：Devu and Flowers

做這個題目出了一些莫名其妙的問題。。

做法：

如果每一個每一種花都有無限個，很明顯，有C（s+n-1,n-1）種取法。

如果某種花取x個以上，那麼就有C（s+n-1-x-1,n-1）種取法。

所以就用到容斥：

#include<stdio.h>#include<iostream>#include<stdlib.h>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<map>using namespace std;#define maxn 21#define LL __int64#define mod 1000000007LL num[21];LL inv[21];void gcd(LL a, LL b, LL& d, LL& x, LL& y) {    if(!b){ d = a; x = 1; y = 0; }    else{ gcd(b, a%b, d, y, x); y -= x*(a/b); }}LL getInv(LL a, LL n) {    LL d, x, y;    gcd(a, n, d, x, y);    return d == 1 ? ( x + n ) % n : -1;}LL com(LL n,LL m){    if(n<m)return 0;    LL ans;    ans=1;    for(LL i=n;i>=(n-m+1);i--)    {        ans=ans*(i%mod)%mod;        ans=ans*inv[n-i+1]%mod;    }    return ans;}int main(){    LL n,s;    for(int i=1;i<=20;i++)inv[i]=getInv(i,mod);    while(~scanf("%I64d%I64d",&n,&s))    {        LL r=0;        int m=n;        for(int i=0;i<m;i++)        {            scanf("%I64d",&num[i]);        }        LL ans=com(s+n-1,n-1);        for(int i=1;i<(1<<m);i++)        {            int x=0;            r=0;            for(int j=0;j<m;j++)            {                if(i&(1<<j))                {                    x++;                    r+=num[j]+1;                }            }            if(x&1)            {                ans-=com(s-r+n-1,n-1);                ans=(ans%mod+mod)%mod;            }            else            {                ans+=com(s-r+n-1,n-1);                ans=(ans%mod+mod)%mod;            }        }        cout<<ans<<endl;    }    return 0;}