More than 2014 schools join-Game 7

1005: (stupid tower defense)

From the question, we obviously know that the Red Tower is the best at the end.

Let's assume that there are j green towers and I-j blue towers in the first one. Therefore, no matter how the sequence of the first I tower is put, the impact on the subsequent tower is exactly the same.

DP [I] [J]: the greatest value of the previous I-tower with J green towers.

DP [I] [J] = max (DP [I-1] [J-1] + energy value obtained from the current tower putting the Green Tower, DP [I-1] [J] + the energy value obtained from the blue tower of the current Tower );

#include <iostream>#include<stdio.h>#include<vector>#include<queue>#include<stack>#include<string.h>#include<algorithm>using namespace std;#define LL __int64#define lcm(a,b) (a*b/gcd(a,b))LL intx(LL num,LL t,LL x,LL chi){    return num*t*(x+chi);}LL inty(LL num,LL t,LL y,LL chi){    return (t)*y*(num-1)*num/2+chi*t*num;}LL intz(LL num,LL t,LL z,LL chi){    return chi*(num*t+(num-1)*num/2*z);}LL dp[1501][1501];int main(){    LL n,x,y,z,t,cas,i,j;    scanf("%I64d",&cas);    LL nn = 1;    while(cas--)    {        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);        LL ans = t*x*n;        LL sum;        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)        {            dp[i][0]=dp[i-1][0];            ans=max(ans,dp[i][0]+intx(n-i,t+i*z,x,0));            for(int j=1;j<=i-1;j++)            {                dp[i][j]=max(dp[i-1][j-1]+inty(1,t+z*(i-j),y,(j-1)*y),                             dp[i-1][j]+intz(1,t+z*(i-1-j),z,j*y));                ans=max(ans,dp[i][j]+intx(n-i,t+(i-j)*z,x,j*y));            }            dp[i][i]=dp[i-1][i-1]+inty(1,t,y,(i-1)*y);            ans=max(ans,dp[i][i]+intx(n-i,t,x,i*y));        }        printf("Case #%I64d: %I64d\n",nn++,ans);    }    return 0;}

1007: (magical forest)

All sorts of discretization, And Then mark the row that the row originally belongs to, then it will be done in a mess.

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <assert.h>#include<iostream>#include<vector>using namespace std;#define MAXN 10050#define LL __int64vector<int>vecx,vecxx;vector<int>vecy,vecyy;struct list{    int x;    int y;    int c;    friend bool operator <(const list &a,const list &b)    {        if(a.x!=b.x)return a.x<b.x;        else return a.y<b.y;    }}node[110000];int fx[110000];int fy[110000];int k;int find(int x,int leap){   // cout<<x<<" "<<leap<<endl;    if(leap==0)    {        int l=0;        int r=vecx.size();        int mid=(l+r)/2;        while(l<r)        {            //cout<<l<<" "<<r<<" "<<mid<<" "<<vecx[mid]<<endl;            if(vecx[mid]<x)l=mid+1;            else r=mid;            mid=(l+r)/2;        }        if(mid!=vecx.size()&&vecx[mid]==x)return mid+1;        else return 0;    }    if(leap==1)    {        int l=0;        int r=vecy.size();        int mid=(l+r)/2;        while(l<r)        {            if(vecy[mid]<x)l=mid+1;            else r=mid;            mid=(l+r)/2;        }        if(mid!=vecy.size()&&vecy[mid]==x)return mid+1;        else return 0;    }    return 0;}void out(int x,int y){  //  cout<<x<<" -- "<<y<<endl;    if(x==0||y==0)    {        cout<<"0"<<endl;        return;    }    int l=1;    int r=k+1;    int mid=(l+r)/2;    while(l<r)    {        if(node[mid].x<x)l=mid+1;        else if(node[mid].x>x)r=mid;        else if(node[mid].x==x)        {            if(node[mid].y<y)l=mid+1;            else if(node[mid].y>=y)r=mid;        }        mid=(l+r)/2;    }    if(node[mid].y==y&&node[mid].x==x&&mid<=k)    {        cout<<node[mid].c<<endl;    }    else cout<<"0"<<endl;}int main(){    int W,cas;    int n,m;    scanf("%d",&W);    cas=0;    while(W--)    {        cas++;        scanf("%d%d%d",&n,&m,&k);        vecx.clear();        vecxx.clear();        vecy.clear();        vecyy.clear();        fx[0]=fy[0]=0;        for(int i=1;i<=k;i++)        {            fx[i]=i;            fy[i]=i;            scanf("%d%d%d",&node[i].x,&node[i].y,&node[i].c);            vecxx.push_back(node[i].x);            vecyy.push_back(node[i].y);        }        sort(vecxx.begin(),vecxx.end());        sort(vecyy.begin(),vecyy.end());        vecx.push_back(vecxx[0]);        vecy.push_back(vecyy[0]);        for(int i=1;i<k;i++)        {            if(vecxx[i]!=vecxx[i-1])vecx.push_back(vecxx[i]);            if(vecyy[i]!=vecyy[i-1])vecy.push_back(vecyy[i]);        }        for(int i=1;i<=k;i++)        {            node[i].x=find(node[i].x,0);            node[i].y=find(node[i].y,1);        }        sort(node+1,node+k+1);        printf("Case #%d:\n",cas);        int q,x,a,b;        scanf("%d",&q);        while(q--)        {            scanf("%d%d%d",&x,&a,&b);            if(x==1)            {                a=find(a,0);                b=find(b,0);                swap(fx[a],fx[b]);            }            if(x==2)            {                a=find(a,1);                b=find(b,1);                swap(fy[a],fy[b]);            }            if(x==3)            {                a=find(a,0);                b=find(b,1);                out(fx[a],fy[b]);            }        }    }    return 0;}

1010: FSF's game

We can know from the question that for any rectangle with N and m length, we can get the score:

Assume that gcd (n, m) = 2 ^ x * 3 ^ y..., and D is a factor of N.

(N, m) = N * m/(gcd (n, m) * (2 ^ 0 + 2 ^ 1 + .. + 2 ^ X) * (3 ^ 0 + 3 ^ 1 +... + 3 ^ y );

Obviously, when n is fixed, M is equal to 1 ~ When n, m/(gcd (n, m) * (2 ^ 0 + 2 ^ 1 + .. + 2 ^ X) * (3 ^ 0 + 3 ^ 1 +... + 3 ^ y) is equal to the sum of all D * (D + 1)/2.

Then we can pre-process the sum of a number factor, and then use the recursive formula directly.

#include <iostream>#include<stdio.h>#include<vector>#include<queue>#include<stack>#include<string.h>#include<algorithm>using namespace std;#define maxn 550000#define UI unsigned int#define N 550000#define LL __int64UI ans[maxn];UI use[maxn];void mk(){    memset(use,0,sizeof(use));    for(UI i=1;i<maxn;i++)    {        LL now=(LL)i*((LL)i+(LL)1)/(LL)2;        UI add=now;        for(UI j=i;j<maxn;j+=i)        {            use[j]=use[j]+add;        }    }}void init(){    mk();    ans[1]=1;    for(UI i=2;i<maxn;i++)    {        ans[i]=ans[i-1]+use[i]*(UI)i;    }}int main(){    int T,n;    init();    scanf("%d",&T);    int cas=0;    while(T--)    {        cas++;        scanf("%d",&n);        printf("Case #%d: %u\n",cas,ans[n]);    }    return 0;}