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LOJ6053简单的函数（min_25筛）
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LOJ6053簡單的函數
https://loj.ac/p/6053
min_25篩模板題，但是要注意質數點有2比較特殊，需要在y==0的時候特判質數包含2的情況。
#include<bits/stdc++.h>
#define LL long long
using namespace std
;
inline LL
read()
{char x
='\0';LL fh
=1,sum
=0;for(x
=getchar();x
<'0'||x
>'9';x
=getchar())if(x
=='-')fh
=-1;for(;x
>='0'&&x
<='9';x
=getchar())sum
=sum
*10+x
-'0';return fh
*sum
;
}
const LL N
=1000009;
const LL mod
=1e9+7;
LL n
,sq
;
LL pri
[N
],pcnt
,sp
[N
];
bool vis
[N
];
inline void init(int mx
)
{vis
[0]=vis
[1]=true;for(LL i
=2;i
<=mx
;i
++){if(!vis
[i
])pri
[++pcnt
]=i
,sp
[pcnt
]=(sp
[pcnt
-1]+i
)%mod
;for(int j
=1;j
<=pcnt
&&i
*pri
[j
]<=mx
;j
++){vis
[i
*pri
[j
]]=true;if(i
%pri
[j
]==0)break;}}
}
LL w
[N
],tot
,ind1
[N
],ind2
[N
],g1
[N
],g0
[N
];
inline LL
S(LL x
,int y
)
{if(x
<=1||pri
[y
]>=x
)return 0;LL k
=(x
<=sq
)?ind1
[x
]:ind2
[n
/x
];LL ans
=g1
[k
]-g0
[k
]-(sp
[y
]-y
);if(y
==0)ans
+=2;for(int i
=y
+1;i
<=pcnt
&&pri
[i
]*pri
[i
]<=x
;i
++){for(LL j
=1,pe
=pri
[i
];pe
<=x
;j
++,pe
*=pri
[i
]){ans
=(ans
+(pri
[i
]^j
)*(S(x
/pe
,i
)+(j
!=1))%mod
)%mod
;}}return ans
;
}
int main()
{n
=read();sq
=sqrt(n
);init(sq
);for(LL lp
=1,rp
=0;lp
<=n
;lp
=rp
+1){rp
=n
/(n
/lp
);w
[++tot
]=n
/lp
;if(w
[tot
]<=sq
)ind1
[w
[tot
]]=tot
;else ind2
[n
/w
[tot
]]=tot
;LL now
=w
[tot
]%mod
;g1
[tot
]=(now
*(now
+1)/2%mod
-1+mod
)%mod
;g0
[tot
]=(now
-1+mod
)%mod
;}
for(int i
=1;i
<=pcnt
;i
++){for(int j
=1;j
<=tot
&&pri
[i
]*pri
[i
]<=w
[j
];j
++){LL now
=w
[j
]/pri
[i
];LL k
=(now
<=sq
)?ind1
[now
]:ind2
[n
/now
];g1
[j
]=(g1
[j
]-pri
[i
]*(g1
[k
]-sp
[i
-1])%mod
+mod
)%mod
;g0
[j
]=(g0
[j
]-(g0
[k
]-(i
-1))+mod
)%mod
;}}
printf("%lld",(S(n
,0)+1)%mod
);return 0;
}
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