1e10内的素数和

板子记录。

  1 #include<bits/stdc++.h>
  2 #define ll long long 
  3 #define scan(i) scanf("%d",&i)
  4 #define scand(i) scanf("%lf",&i)
  5 #define scanl(i) scanf("%lld",&i)
  6 #define f(i,a,b) for(int i=a;i<=b;i++) 
  7 #define pb(i) push_back(i)
  8 #define ppb pop_back()
  9 #define pf printf
 10 #define dbg(args...) cout<<#args<<" : "<<args<<endl;
 11 #define input freopen("in.txt","r",stdin)
 12 #define output freopen("out.txt","w",stdout)
 13 #define io ios::sync_with_stdio(0)
 14 
 15 #define LINT long long
 16 using namespace std;
 17 
 18 inline LINT V2IDX(LINT v, LINT N, LINT Ndr, LINT nv) {
 19     return v >= Ndr ? (N/v - 1) : (nv - v);
 20 }
 21 
 22 LINT primesum(LINT N) {
 23     LINT *S;
 24     LINT *V;
 25 
 26     LINT r = (LINT)sqrt(N);
 27     LINT Ndr = N/r;
 28 
 29 //    assert(r*r <= N and (r+1)*(r+1) > N);
 30 
 31     LINT nv = r + Ndr - 1;
 32 
 33     V = new LINT[nv];
 34     S = new LINT[nv];
 35 
 36     for (LINT i=0; i<r; i++) {
 37         V[i] = N/(i+1);
 38     }
 39     for (LINT i=r; i<nv; i++) {
 40         V[i] = V[i-1] - 1;
 41     }
 42 
 43     for (LINT i=0; i<nv; i++) {
 44         S[i] = V[i] * (V[i] + 1) / 2 - 1;
 45     }
 46 
 47     for (LINT p=2; p<=r; p++) {
 48         if (S[nv-p] > S[nv-p+1]) {
 49             LINT sp = S[nv-p+1];
 50             LINT p2 = p*p;
 51             for (LINT i=0; i<nv; i++) {
 52                 if (V[i] >= p2) {
 53                     S[i] -= p * (S[V2IDX(V[i]/p, N, Ndr, nv)] - sp);
 54                 } else {
 55                     break;
 56                 }
 57             }
 58         }
 59     }
 60     return S[0];
 61 }
 62 
 63 ostream& operator<<(ostream& os, __int128 t) {
 64     if (t==0) return os << "0";
 65     if (t<0) {
 66         os<<"-";
 67         t=-t;
 68     }
 69     int a[50],ai=0;
 70     memset(a,0,sizeof a);
 71     while (t!=0){
 72         a[ai++]=t%10;
 73         t/=10;
 74     }
 75     for (int i=1;i<=ai;i++) os<<abs(a[ai-i]);
 76     return os<<"";
 77 }
 78 
 79 int main() {
 80 //    LINT N = 1000000000;
 81 //    printf("%lld\n", primesum(N));
 82 //cout<<primesum(5)<<endl;
 83 //freopen("out.txt","r",stdin);
 84     io;
 85     int t;
 86     ll mod,n;
 87     scan(t);
 88     f(kk,1,t){
 89         scanl(n);
 90         scanl(mod);
 91         if(n==1){
 92             puts("0");
 93         }
 94         else if(n==2){
 95             puts("6");
 96         }
 97         else{
 98             __int128 ans=__int128(n+2)*__int128(n+1)/2;
 99             ans%=mod;
100             ans+=primesum(n+1);
101             ans-=5;
102             ans=ans%mod;
103 //            cout<<ans<<endl;
104             ll x=ans;
105             pf("%lld\n",x);
106         }
107 //        ll ans2=ans%mod;
108 //        pf("%lld\n",ans);
109     }
110 }

 

posted @ 2020-10-09 16:56  Lovaer  阅读(267)  评论(0)  编辑  收藏  举报