CQOI2015 解题报告

CQOI2015终于全做完了~~~，讲一下题吧

首先这套题比起其他省选还是比较水的，就是5道题比较蛋疼

T1：[CQOI2015]选数

这道题还是比较神的。

首先给个比较神的题解：popoqqq大神的blog这个莫比乌斯反演真的不会

我们记f[i]为gcd=i*k时的个数，可以得到若数都不相等的话，i一定小于1e5（辗转相减法可得），那么当数都不相等时，答案显然为(r/(k*i)-l/(k*i)+1)^n-(r/(k*i)-l/(k*i)+1)-sigma(f[i*j])然后就能愉快的推出来啦，还有就是当l=1时要特判一下

CODE：

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 #define mod 1000000007
 typedef long long ll;
 int power(ll x,ll y) {
     ll ans=;
     for (;y;y>>=){
         if (y&) (ans*=x)%=mod;
         (x*=x)%=mod;
     }
     return ans;
 }
 ll n,l,k,h;
 #define maxk 100000
 ll f[maxk+];
 int main(){
     freopen("number.in","r",stdin);
     freopen("number.out","w",stdout);
     ll L,H;
     scanf("%lld%lld%lld%lld",&n,&k,&L,&H);
     ll l=L/k,h=H/k;
     if (L%k) l++;
     int tmp=;
     for (int i=maxk;i;i--) {
         ll L=l/i,H=h/i;
         if (l%i) L++;
         f[i]=(power(H-L+,n)-(H-L+)+mod)%mod;
         for (int j=i+i;j<=maxk;j+=i) f[i]=(f[i]-f[j]+mod)%mod;
     }
     if (l==) f[]++;
     printf("%lld\n",f[]);
     return ;
 }

T2:[CQOI2015]网络吞吐量

这不就是先求出最短路图后拆点跑个最大流么= =。直接做就行了不要考虑时间问题。。。

CODE：

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 using namespace std;
 typedef long long ll;
 #define maxm 150000
 #define maxn 1100
 struct edges{
     int to,next,dist;ll cap;
 }edge[maxm*];
 int l,next[maxn];
 inline void _addedge(int x,int y,int z){
     l++;
     edge[l*]=(edges) {x,next[y],z,};next[y]=l*;
     edge[l*+]=(edges) {y,next[x],z,};next[x]=l*+;
 }
 inline void addedge(int x,int y,ll z){
     l++;
     edge[l*]=(edges){y,next[x],,z};next[x]=l*;
     edge[l*+]=(edges){x,next[y],,};next[y]=l*+;
 }
 ll dist[maxn];
 bool b[maxn];
 typedef pair<ll,int> ii;
 priority_queue<ii,vector<ii>,greater<ii> > q;
 #define fi first
 #define se second
 #define inf 0x7fffffff
 #define Inf inf*1ll*inf
 int n;
 inline void dij(){
     for (int i=;i<=n;i++) dist[i]=Inf;
     dist[]=;
     q.push(ii(,));
     while (!q.empty()){
         ii u=q.top();q.pop();
         if (b[u.se]) continue;
         b[u.se]=;
         for (int i=next[u.se];i;i=edge[i].next)
             if (dist[u.se]+edge[i].dist<dist[edge[i].to]) {
                 dist[edge[i].to]=dist[u.se]+edge[i].dist;
                 q.push(ii(dist[edge[i].to],edge[i].to));
             }
     }
 }
 int p[maxn],gap[maxn],h[maxn],s,t,N;
 ll sap(int u,ll flow){
     if (u==t) return flow;
     ll cnt=;
     for (int i=p[u];i;i=edge[i].next)
         if (edge[i].cap&&h[edge[i].to]+==h[u]) {
             ll cur=sap(edge[i].to,min(edge[i].cap,flow-cnt));
             edge[i].cap-=cur;edge[i^].cap+=cur;
             p[u]=i;
             if ((cnt+=cur)==flow) return flow;
         }
     if (!(--gap[h[u]])) h[s]=N;
     gap[++h[u]]++;
     p[u]=next[u];
     return cnt;
 }
 inline ll maxflow(){
     for (int i=;i<=N;i++) p[i]=next[i];
     ll flow=;
     gap[]=N;
     while (h[s]<N) flow+=sap(s,Inf);
     return flow;
 }
 int main(){
     freopen("network.in","r",stdin);
     freopen("network.out","w",stdout);
     int m;
     scanf("%d%d",&n,&m);
     N=n*;
     for (int i=;i<=m;i++) {
         int x,y,z;
         scanf("%d%d%d",&x,&y,&z);
         _addedge(x,y,z);
     }
     dij();
     memset(next,,sizeof(next));
     l=;
     for (int i=;i<=m*+;i+=) {
         if (dist[edge[i].to]==dist[edge[i^].to]+edge[i].dist)
             addedge(n+edge[i^].to,edge[i].to,Inf);
         else if (dist[edge[i^].to]==dist[edge[i].to]+edge[i^].dist)
             addedge(n+edge[i].to,edge[i^].to,Inf);
     }
     for (int i=;i<=n;i++) {
         int x;
         scanf("%d",&x);
         addedge(i,n+i,x);
     }
     s=+n,t=n;
     printf("%lld\n",maxflow());
     return ;
 }

T3：[CQOI2015]任务查询系统

描述：戳我~~~

这道题首先很明显是道裸的数据结构题啦。首先他要求在线，那么按顺序建个函数式线段树就行啦

自己太弱调了好久= =

CODE：

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 using namespace std;
 typedef long long ll;
 #define pb push_back
 #define maxn 100100
 struct qnode{
     int x,y,z;
 }q[maxn*];
 int l;
 inline void add(int x,int y,int z) {q[++l]=(qnode){x,y,z};}
 bool cmp(qnode x,qnode y) {return x.x<y.x;}
 struct node{int lc,rc,sum;ll s;}t[maxn*];
 #define mid (l+r>>1)
 int cnt=;
 int ins(int x,int l,int r,int y,int z){
     int root=++cnt;
     t[root]=t[x];
     t[root].s+=y*z;
     t[root].sum+=z;
     if (l==r) return root;
     if (y<=mid) t[root].lc=ins(t[x].lc,l,mid,y,z);
     else t[root].rc=ins(t[x].rc,mid+,r,y,z);
     return root;
 }
 ll que(int x,int l,int r,int &k){
     if (t[x].sum<=k) {
         k-=t[x].sum;return t[x].s;
     }
     if (l==r) {ll ans=l*1ll*k;k=;return ans;}
     ll ans=que(t[x].lc,l,mid,k);
     if (k) ans+=que(t[x].rc,mid+,r,k);
     return ans;
 }
 vector<int> a;
 int s[maxn],e[maxn],p[maxn],root[maxn];
 int main(){
     int n,m;
     scanf("%d%d",&n,&m);
     a.pb();
     for (int i=;i<=n;i++) {
         scanf("%d%d%d",s+i,e+i,p+i);
         e[i]++;
         a.pb(s[i]);a.pb(e[i]);
     }
     sort(a.begin(),a.end());
     a.resize(unique(a.begin(),a.end())-a.begin());
     for (int i=;i<=n;i++) {
         add(lower_bound(a.begin(),a.end(),s[i])-a.begin(),p[i],);
         add(lower_bound(a.begin(),a.end(),e[i])-a.begin(),p[i],-);
     }
     sort(q+,q++l,cmp);
     #define inf 10000000
     for (int i=;i<=l;i++) root[q[i].x]=ins(root[q[i-].x],,inf,q[i].y,q[i].z);
     ll pre=;
     for (int i=;i<=m;i++) {
         int x,ai,bi,ci;
         scanf("%d%d%d%d\n",&x,&ai,&bi,&ci);
         int k=1ll+(ai*1ll*(pre%ci)%ci+bi)%ci;
         pre=que(root[upper_bound(a.begin(),a.end(),x)-a.begin()-],,inf,k);
         printf("%lld\n",pre);
     }
     return ;
 }

T4：[CQOI2015]多项式

描述：戳我~~~

又是高精度= =

首先我们可以换一下元，吧x-t换成x那就能得到

那么就能算5次即可啦，再考虑怎么快速算组合数

完成啦

CODE：

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #define maxb 4
 using namespace std;
 char s[];
 int p[]={,,,,,,,,};
 struct bignum{
     int a[],n,flag;
     bignum(int x){
         n=;memset(a,,sizeof(a));
         flag=x<?:;x=abs(x);
         while (x) {a[++n]=x%p[maxb];x/=p[maxb];}
         if (!n) n=;
     }
     bignum(){n=;memset(a,,sizeof(a));flag=;}
     int read(){
         scanf("%s",s);
         int len=strlen(s);
         n=;
         if (len==&&s[]=='') {a[n=]=;return ;}
         for (int i=len;i>=;i-=maxb) {
             int t=;
             if (i>maxb) for (int j=i-maxb+;j<=i;j++) t=t*+s[j-]-'';
             else for (int j=;j<=i;j++) t=t*+s[j-]-'';
             a[++n]=t;
         }
         return ;
     }
     int write(){
         if (flag) printf("-");
         printf("%d",a[n]);
         for (int i=n-;i>=;i--) {
             for (int j=;j<maxb;j++) if  (a[i]<p[j]) printf("");
             printf("%d",a[i]);
         }
         printf("\n");
         return ;
     }
     int cmp(bignum x){
         if (x.n!=n) return (x.n<n);
         for (int i=n;i>=;i--)
             if (x.a[i]!=a[i]) return (x.a[i]<a[i]);
         return -;
     }
     int div(int x){
         int s=;
         for (int i=n;i>=;i--){
             s=a[i]+s*p[maxb];
             a[i]=s/x;
             s%=x;
         }
         if (n!=&&!a[n]) n--;
         return s;
     }
 }a,b;
 bignum operator + (bignum x,bignum y){
     if (x.flag^y.flag && x.cmp(y)==)swap(x,y);
     if (x.flag==y.flag) {
         x.n=max(x.n,y.n);
         for (int i=;i<=x.n;i++) {
             x.a[i]+=y.a[i];
             x.a[i+]+=x.a[i]/p[maxb];
             x.a[i]%=p[maxb];
         }
         if (x.a[x.n+]) x.n++;
         return x;
     }
     for (int i=;i<=x.n;i++) {
         x.a[i]-=y.a[i];
         if (x.a[i]<) {x.a[i]+=p[maxb];x.a[i+]--;}
         while (!x.a[x.n]&&x.n>) x.n--;
     }
     return x;
 }
 bignum operator * (bignum x,bignum y){
     bignum ans;
     if (x.n==&&x.a[]==) return ans;
     if (y.n==&&y.a[]==) return ans;
     ans.flag=x.flag^y.flag;
     for (int i=;i<=x.n;i++)
         for (int j=;j<=y.n;j++) {
             ans.a[i+j-]+=x.a[i]*y.a[j];
             ans.a[i+j]+=ans.a[i+j-]/p[maxb];
             ans.a[i+j-]%=p[maxb];
         }
     int n=x.n+y.n-;
     while (ans.a[n+]) {
         n++;ans.a[n+]+=ans.a[n]/p[maxb];
         ans.a[n]%=p[maxb];
     }
     ans.n=n;
     return ans;
 }
 int sum;
 int main(){
     static bignum n,m;
     int t;
     n.read();
     scanf("%d",&t);
     m.read();
     static int a[];
     a[]=;
     for (int i=;i<=;i++) a[i]=(a[i-]*+) % ;
     static bignum tmp=m;
     int l=tmp.div();
     static bignum ans,c=bignum(),T=bignum();
     int cnt=;
     for (bignum i=m;i.cmp(n)!=;i=i+bignum(),cnt++) {
         ans=ans+c*T*bignum(a[l]);
         l=(l+)%;
         c=c*(i+bignum());
         c.div(cnt+);
         T=T*bignum(t);
     }
     ans.write();
     return ;
 }

T5:[CQOI2015]标示设计

描述：戳我~~~

首先我们考虑轮廓线，那么每个L可能有四种状态：已经结束，还未开始，还没拐弯，已经拐弯，那么如果从上到下，从左到右做的话，可以发现最多只有3条横边被标记，那么对每个L我们可以用一维来存这个状态，在加上起始状态还有终止状态就行啦，然后已经拐弯的状态一定是在那条竖边上在讨论下就行啦

由于我是从左到右，从上到下做的，所以每一维就多了1倍的状态，所以在BZ上就光荣的TLE啦，不过在学校的oj上还是能过的。所以代码就不贴啦