結果

問題 No.1661 Sum is Prime (Hard Version)
ユーザー LayCurse
提出日時 2021-07-29 21:23:46
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,565 ms / 3,000 ms
コード長 1,705 bytes
コンパイル時間 1,878 ms
コンパイル使用メモリ 193,980 KB
最終ジャッジ日時 2025-01-23 10:12:45
ジャッジサーバーID
(参考情報)
judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 22
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:50:8: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   50 |   scanf("%lld%lld",&L,&R);
      |   ~~~~~^~~~~~~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int ps, p[1000000];
ll memo[] = {16252325,31324703,46009215,60454705,74726528,88862422,102886526,116818447,130670192,144449537,158165829,171827136,185436625,198996103
    ,212514323,225991743,239429993,252834065,266206294,279545368,292856421,306137611,319391721,332620900,345826612,359006517,372168766,385307831
    ,398425675,411523195,424603409,437663672,450708777,463733626,476741968,489736021,502712575,515673696,528624525,541555851,554476170,567382703
    ,580275482,593155089,606022680,618878615,631717982,644550922,657371139,670180516,682979568,695766925,708543557,721310048,734065508,746813071
    ,759550004,772276773,784994469,797703398,810401956,823092766,835773764,848450250,861116918,873772692};
bitset<500000> b;
ll solve(ll n){
int i;
ll s, k, res;
if(n < 1000000){
for(i=0;i<ps;i++) if(p[i] > n) break;
return i;
}
s = n / 300000000;
if(s){
res = memo[s-1];
s *= 300000000;
} else {
res = ps;
s = 1000000;
}
for(;;s+=1000000){
b.set();
for(i=1;;i++){
k = (ll) p[i] * p[i] - s;
if(k >= 1000000) break;
if(k < 0) k += ((-k+p[i]-1) / p[i]) * p[i];
if(k % 2 == 0) k += p[i];
k /= 2;
while(k < 500000) b.reset(k), k += p[i];
}
if(n-s < 1000000){
for(i=1;i<=n-s;i+=2) res += b[i/2];
break;
}
res += b.count();
}
return res;
}
int main(){
int i, j;
ll L, R;
scanf("%lld%lld",&L,&R);
p[2] = 1;
for(i=3;i<1000000;i+=2) p[i] = 1;
for(i=3;i<1000;i+=2) if(p[i]) for(j=i*i;j<1000000;j+=i) p[j] = 0;
for(i=0;i<1000000;i++) if(p[i]) p[ps++] = i;
printf("%lld\n", solve(2*R) - solve(2*L) + solve(R) - solve(L-1));
}
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