結果
問題 | No.1661 Sum is Prime (Hard Version) |
ユーザー |
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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 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 22 |
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
#include<bits/stdc++.h>using namespace std;#define ll long longint 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));}