bzoj3529: [Sdoi2014]数表

来源：互联网发布：小约翰震惊知乎编辑：程序博客网时间：2024/05/16 11:28

链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3529

题意：中文题。。。

分析：同bzoj2301，论文题。。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<math.h>#include<cstdio>#include<vector>#include<string>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=100010;const int MAX=151;const int MOD=1000000007;const int MOD1=100000007;const int MOD2=100000009;const int INF=2100000000;const double EPS=0.00000001;typedef long long ll;typedef unsigned long long uI64;int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}struct node1 {    int b,n,m,x;}q[20010];int cmd1(node1 a,node1 b) {    return a.x<b.x;}struct node2 {    int x,y;}f[N];int cmd2(node2 a,node2 b) {    return a.y<b.y;}int a[N],bo[N],mu[N],ans[N],sum[N];void deal() {    int i,j,k,w,n=100000;    memset(bo,0,sizeof(bo));    f[1].x=f[1].y=mu[1]=1;k=0;    for (i=2;i<=n;i++) {        f[i].x=i;        if (!bo[i]) { a[++k]=i;f[i].y=i+1;mu[i]=-1; }        for (j=1;j<=k;j++) {            if (a[j]*i>n) break ;            bo[a[j]*i]=1;f[a[j]*i].y=f[i].y*(a[j]+1);            if (i%a[j]==0) {                f[a[j]*i].y-=f[i/a[j]].y*a[j];                mu[a[j]*i]=0;break ;            }            mu[a[j]*i]=-mu[i];        }    }    sort(f+1,f+n+1,cmd2);}void add(int x,int y) {    for (;x<=100000;x+=x&(-x)) sum[x]+=y;}void addup(int a) {    int w=f[a].x;    while (w<=100000) { add(w,mu[w/f[a].x]*f[a].y);w+=f[a].x; }}int getsum(int x) {    int ret=0;    for (;x;x-=x&(-x)) ret+=sum[x];    return ret;}int getans(int n,int m) {    int i,last,ret=0;    if (n>m) { n^=m;m^=n;n^=m; }    for (i=1;i<=n;i=last+1) {        last=min(n/(n/i),m/(m/i));        ret+=(getsum(last)-getsum(i-1))*(n/i)*(m/i);    }    return ret;}int main(){    deal();    int i,k,t;    scanf("%d", &t);    for (i=1;i<=t;i++) {        q[i].b=i;        scanf("%d%d%d", &q[i].n, &q[i].m, &q[i].x);    }    sort(q+1,q+t+1,cmd1);    k=0;    memset(sum,0,sizeof(sum));    memset(ans,0,sizeof(ans));    for (i=1;i<=t;i++) {        while (k<100000&&f[k+1].y<=q[i].x) addup(++k);        ans[q[i].b]=getans(q[i].n,q[i].m);    }    for (i=1;i<=t;i++) printf("%d\n", ans[i]&0x7fffffff);    return 0;}/*44 4 310 10 5100 100 10100000 100000 7*/

0 0