結果
| 問題 |
No.1661 Sum is Prime (Hard Version)
|
| コンテスト | |
| ユーザー |
LayCurse
|
| 提出日時 | 2021-07-29 21:28:04 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 1,156 ms / 3,000 ms |
| コード長 | 2,041 bytes |
| コンパイル時間 | 2,397 ms |
| コンパイル使用メモリ | 193,700 KB |
| 最終ジャッジ日時 | 2025-01-23 10:13:22 |
|
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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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 long
int ps, p[1000000];
ll memo[] = {11078937,21336326,31324703,41146179,50847534,60454705,69985473,79451833,88862422,98222287,107540122,116818447,126062167,135270258,144449537,153600805,162725196,171827136,180906194,189961812,198996103,208013454,217011319,225991743,234954223,243902342,252834065,261751864,270655552,279545368,288422869,297285198,306137611,314977166,323804352,332620900,341426904,350221825,359006517,367783654,376549859,385307831,394055910,402793457,411523195,420243162,428958595,437663672,446362736,455052511,463733626,472408200,481074475,489736021,498388617,507036251,515673696,524309392,532936342,541555851,550170437,558778993,567382703,575978253,584570200,593155089,601735269,610308664,618878615,627440336,635997249,644550922,653099304,661643304,670180516,678714823,687242934,695766925,704286233,712799821,721310048,729813991,738315156,746813071,755305935,763794029,772276773,780756650,789230673,797703398,806169530,814633249,823092766,831548431,840000027,848450250,856895823,865335133,873772692,882206716};
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 / 200000000;
if(s){
res = memo[s-1];
s *= 200000000;
} 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));
}
LayCurse