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

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

A:Game With Sticks

水题。。。每次操作，都会拿走一个横行，一个竖行。

所以一共会操作min（横行，竖行）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

给数组中的每一个数字标号，标记他应该出现在哪一个位置。

然后从头往后找，如果当前位置的数字不是应该在这个位置的数字。

那么就找到应该在这个位置的数字的位置，然后翻转。如果还不是按顺序排的话，就输出no，否则输出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

枚举两个差值的正负号。

对于每一种情况：

可以算出A，B,C最少赢几个球。然后看看当前的赢球数是不是符合K。

然后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

假如字符串为：abbabbbaaa

我们把这个字符串如下记录: 

     字符串： a b a b a

数组num ：1 2 1 3  3

数组的每一项代表这个字符串这个位置的字符是由几个字符压缩成的。

对于回文串的两边都是a的情况：

预处理从起点走奇数步可到达多少a，走偶数步，可到达多少a。

然后从第一个a往后走，可在O（1）的复杂度内得出当前a的为回文串的左边，一共有几个奇数子串，几个偶数子串。

对于回文串的两边都是b的情况，类似与两边都是a的情况。

#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

做这个题目出了一些莫名其妙的问题。。

做法：

如果每一个每一种花都有无限个，很明显，有C（s+n-1,n-1）种取法。

如果某种花取x个以上，那么就有C（s+n-1-x-1,n-1）种取法。

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;}