151224总结

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151223

TJOI2015D1

T1 有意义的字符串

然后就可以矩阵快速幂求出a_n+1啦~

注意， 1.n=0时要特判，就是在这儿丢了5分。。。

2.模数非常大，必须开unsigned long long ，然后写快速乘

复杂度：O(2³logNlogx)

需要的知识：二阶递推、矩阵乘法

T2 城池攻占

可并堆，从叶子往根合并，当堆顶值不满足时就弹出，每个点只加入一次、弹出一次。在树上每个节点改变值的时候，像双标记线段树一样，打lazy标记。

卡常。。。

复杂度 O(NlogN)

需要的知识：可并堆

T3 装备购买

复杂度：O(N³)

需要的知识：贪心、高斯消元

#include<iostream>#include<algorithm>#include<cstdlib>#include<cstdio>#include<string>#include<cstring>#include<ctime>#include<cmath>#define LL long long#define ul unsigned long longusing namespace std;const ul M = 7528443412579576937;LL b,d,n,x,y;ul ret[2][2],a[2][2],t[2][2];ul ksc(ul a,ul b){ul ret=0;for (;b;b>>=1,a=(a<<1)%M)if (b&1) ret=(ret+a)%M;return ret;}void mul(ul a[][2],ul b[][2]){for (int i=0;i<2;i++)for (int j=0;j<2;j++){t[i][j]=0;for (int k=0;k<2;k++)t[i][j]=(t[i][j]+ksc(a[i][k],b[k][j]))%M;}for (int i=0;i<2;i++)for (int j=0;j<2;j++)a[i][j]=t[i][j];}void ksm(LL b){ret[0][0]=ret[1][1]=a[1][0]=1;for (;b;b>>=1,mul(a,a))if (b&1) mul(ret,a);}int main(){scanf("%I64d%I64d%I64d",&b,&d,&n);if (!n) {puts("1");return 0;}a[0][0]=b;a[0][1]=(d-b*b)>>2;ksm(n-1);ul ans=((ret[0][1]<<1)%M+ksc(ret[0][0],b))%M;if (b*b!=d&&(~n&1)) ans--;if (ans<0) ans+=M;cout<<ans<<endl;return 0;}

#include<iostream>#include<algorithm>#include<cstdlib>#include<cstdio>#include<string>#include<cstring>#include<ctime>#include<cmath>#define N 300005#define LL long longusing namespace std;int n,m;int first[N],next[N];int root[N],father[N];LL h[N],a[N],v[N];LL w[N],d[N],lc[N],rc[N],add[N],mul[N],as[N];int p[N];int ans1[N],ans2[N];LL get_int(){LL ret=0;char c=getchar();bool t=0;for (;(c<'0'||c>'9')&&c!='-';c=getchar());if (c=='-') c=getchar(),t=1;for (;c>='0'&&c<='9';c=getchar())ret=(ret<<1)+(ret<<3)+c-'0';return t ? -ret : ret;}void push_down(int x){int l=lc[x],r=rc[x];if (as[x]) {ans2[l]+=as[x],ans2[r]+=as[x];as[l]+=as[x],as[r]+=as[x],as[x]=0;}if (mul[x]^1) {w[l]*=mul[x],w[r]*=mul[x];mul[l]*=mul[x],mul[r]*=mul[x];add[l]*=mul[x],add[r]*=mul[x];mul[x]=1;}if (add[x]) {w[l]+=add[x],w[r]+=add[x];add[l]+=add[x],add[r]+=add[x];add[x]=0;}}int merge(int x,int y){if (!x) return y;if (!y) return x;if (w[x]>w[y]) swap(x,y);push_down(x);rc[x]=merge(rc[x],y);if (d[lc[x]]<d[rc[x]])swap(lc[x],rc[x]);d[x]=d[rc[x]]+1;return x;}void init(){n=get_int();m=get_int();fill(mul+1,mul+m+1,1);for (int i=1;i<=n;i++) h[i]=get_int();for (int i=2;i<=n;i++){father[i]=get_int();a[i]=get_int();v[i]=get_int();next[i]=first[father[i]];first[father[i]]=i;}int x;for (int i=1;i<=m;i++){w[i]=get_int();x=get_int();root[x]=merge(root[x],i);}}void work(){for (int x=n;x;x--){int rt=root[x];for (int i=first[x];i;i=next[i])rt=merge(rt,root[i]);while (rt&&w[rt]<h[x]){push_down(rt);rt=merge(lc[rt],rc[rt]),ans1[x]++;}root[x]=rt;if (rt){as[rt]++,ans2[rt]++;if (a[x]) w[rt]*=v[x],add[rt]*=v[x],mul[rt]*=v[x];else w[rt]+=v[x],add[rt]+=v[x];}}if (root[1]){int head=0,tail=1;p[1]=root[1];while (head^tail){int x=p[++head];push_down(x);if (lc[x]) p[++tail]=lc[x];if (rc[x]) p[++tail]=rc[x];}}}int main(){init();work();for (int i=1;i<=n;i++) printf("%d\n",ans1[i]);for (int i=1;i<=m;i++) printf("%d\n",ans2[i]);return 0;}

#include<iostream>#include<algorithm>#include<cstdlib>#include<cstdio>#include<string>#include<cstring>#include<ctime>#include<cmath>#define D double#define eps 0.00001#define N 505using namespace std;int n,m;int w[N];struct node{D z[N];int c; }a[N];bool cmp(const node &a,const node &b){return a.c<b.c;}void init(){scanf("%d%d",&n,&m);for (int i=1;i<=n;i++)for (int j=1;j<=m;j++)scanf("%lf",&a[i].z[j]);for (int i=1;i<=n;i++) scanf("%d",&a[i].c);sort(a+1,a+n+1,cmp);}void work(){int num=0,ans=0;for (int i=1;i<=n;i++)for (int j=1;j<=m;j++)if (fabs(a[i].z[j])>eps){if (w[j]){D t=a[i].z[j]/a[w[j]].z[j];for (int k=j;k<=m;k++)a[i].z[k]-=t*a[w[j]].z[k];}else{w[j]=i;num++;ans+=a[i].c;break;}}printf("%d %d\n",num,ans);}int main(){init();work();return 0;}

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