BZOJ 3813: 奇数国【水题】

来源：互联网发布：百富环球软件编辑：程序博客网时间：2024/05/13 04:21

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3813

就是求欧拉函数啦，利用积性函数性质将数字按质因数分解，开60个树状数组搞一搞就是啦。【因为只保存指数所以用树状数组】

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int MOD = 19961993;int n,m,cnt = 0,sum[61][100001],P[61][100001],p[100+10];bool flag[1000+10];void add(int T[],int x,int w) {while(x <= n) {T[x] += w;x += (x&-x);}}int query(int T[],int x) {int ret = 0;while(x) {ret += T[x];x -= (x&-x);}return ret;}void init() {for(int i = 2; i <= 281 ; i++) {if(!flag[i])p[++cnt] = i;for(int j = 1; j <= cnt && i * p[j] <= 281; j++) {flag[i * p[j]] = 1;if(i % p[j] == 0)break;}}for(int i = 1; i <= n; i++){P[2][i] = 1;add(sum[2],i,1);}}int mul(int di,int x) {int ret = 1;while(x) {if(x&1)ret = (1LL * ret * di) % MOD;x >>= 1;di = (1LL * di * di) % MOD;}return ret;}int query(int l,int r) {int ans = 1;bool flag = true;for(int i = 1; i <= cnt; i++) {int ret = query(sum[i],r)-query(sum[i],l-1);if(ret)flag = false;if(ret)ans = (1LL * ans * (p[i]-1) * mul(p[i],ret-1)) % MOD;}if(flag)return 0;return ans;}void change(int po,int f) {for(int i = 1; i <= cnt; i++) {if(P[i][po])add(sum[i],po,-P[i][po]);P[i][po] = 0;}for(int i = 1; i <= cnt; i++) {int ret = 0;while(f%p[i] == 0) {ret++;f /= p[i];}if(ret)add(sum[i],po,ret);P[i][po] = ret;}}int main() {n = 100000;int a,b,c;init();scanf("%d",&m);for(int i = 1; i <= m; i++) {scanf("%d%d%d",&a,&b,&c);if(a == 0) {printf("%d\n",query(b,c));}else {change(b,c);}}return 0;}

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