2017 Multi-University Training Contest 4 1003
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题意:让你求【L,R】区间的每个数i的k次方的因子数和%998244353。
解法:首先要知道一个数的因子数,是等于由这个质因子的次方+1相乘起来等到的。设n=p1c1p2c2...pmcm,则d(nk)=(kc1+1)(kc2+1)...(kcm+1)。如果是k次方的话就把这个k乘到质因子的次方上去就行了,关键在于怎么枚举区间,首先枚举小于等于根号R区间的质数p,再枚举【L,R】区间所有p的倍数,计算他的因子数,剩下的就是超过根号R的质数,只可能为0或者1。
#include <stdio.h>#include<math.h>#include<cstring>#include<iostream>#define LL long longconst LL mod = 998244353;const int MAX_N = 1e6+10;using namespace std;LL l,r,k,ans[MAX_N],f[MAX_N];int isp[MAX_N],cnt;LL pri[MAX_N];void Euler(LL p){ LL t = 0; for(LL i = l/p*p;i <= r;i+=p) { if(i >= l) { t = 0; while(f[i-l]%p==0) { t++; f[i-l]/=p; } ans[i-l] = ans[i-l]*(t*k+1)%mod; } }}void init(){ cnt = 0; memset(isp,0,sizeof(isp)); isp[0] = 1; isp[1] = 1; for(int i = 2;i < MAX_N;i++) { if(!isp[i]) pri[cnt++] = i; for(int j = 0;j < cnt&&pri[j]*i < MAX_N;j++) { isp[i*pri[j]] = 1; if(i%pri[j] == 0) break; } }}int main(){ int T; init(); //cout<<cnt<<endl; scanf("%d",&T); while(T--) { scanf("%lld%lld%lld",&l,&r,&k); LL d = r-l; for(int i = 0;i <= d;i++) { f[i] = i+l; ans[i] = 1; } for(int i = 0;i < cnt;i++) { if(pri[i]*pri[i] > r) break; Euler(pri[i]); } LL ANS = 0; for(int i = 0;i <= d;i++) { if(f[i] > 1) ans[i] = ans[i]*(k+1)%mod; ANS = (ANS+ans[i])%mod; } printf("%lld\n",ANS); } return 0;}
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