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P4619 [SDOI2018]旧试题
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P4619 [SDOI2018]旧试题
题意：
求个式子：
(∑i=1A∑j=1B∑k=1Cd(i∗j∗k))mod(109+7)(\sum_{i=1}^{A}\sum_{j=1}^{B}\sum_{k=1}^{C}d(i*j*k))mod(10^9+7)(i=1∑Aj=1∑Bk=1∑Cd(i∗j∗k))mod(109+7)
题解：
原创博文1k纪念
很明显，莫比乌斯反演
马上更新
三元环链接
此等好题，作为第1000题的纪念
代码：
这个题让我写，我是真写不出来，非常适合练练莫比乌斯推导
代码条例非常清晰，很适合阅读
#include <bits/stdc++.h>
#define ll long long
using namespace std
;
const int N
= 1e5 + 7;
const int M
= 1e6 + 7;
const int mod
= 1e9 + 7;
bool Prime
[N
];
int tot
, P
[N
>> 3], mu
[N
];
void get_mu()
{mu
[1]= 1;for (int i
= 2; i
<= 1e5; i
++) {if (!Prime
[i
])P
[++tot
]= i
, mu
[i
]= -1;for (int j
= 1; P
[j
] * i
<= 1e5 && j
<= tot
; j
++) {Prime
[P
[j
] * i
]= 1;if (i
% P
[j
] == 0) {mu
[P
[j
] * i
]= 0;break;}mu
[P
[j
] * i
]= -mu
[i
];}}
}
vector
<int> ve
[N
];
unordered_map
<int, bool> mp
[N
];
int T
, Max
, A
, B
, C
, ans
, p
[N
];
int sA
[M
], sB
[M
], sC
[M
], cnt
, deg
[N
];
int gcd(int x
, int y
)
{return x
== 0 ? y
: gcd(y
% x
, x
);
}
ll
lcm(int x
, int y
)
{return 1ll * x
/ gcd(x
, y
) * y
;
}
int ta
, tb
, tc
, sum
;
void cal(int a
, int b
, int c
)
{sum
=(sum
+ 1ll * p
[a
/ ta
] * p
[b
/ tb
] % mod
* p
[c
/ tc
] % mod
)% mod
;
}
void get(int a
, int b
, int c
)
{sum
= 0;if (a
== b
&& b
== c
)cal(A
, B
, C
);else if (a
== b
|| b
== c
|| c
== a
) {cal(A
, B
, C
);cal(C
, A
, B
);cal(B
, C
, A
);}else {cal(A
, B
, C
);cal(A
, C
, B
);cal(B
, A
, C
);cal(B
, C
, A
);cal(C
, A
, B
);cal(C
, B
, A
);}ans
= (ans
+ (mu
[a
] * mu
[b
] * mu
[c
] * sum
% mod
+ mod
) % mod
) % mod
;
}
int vis
[N
];
void solve()
{for (int i
= 1; i
<= cnt
; i
++) { int &u
= sA
[i
], &v
= sB
[i
];if (deg
[u
] < deg
[v
])swap(u
, v
);else if (deg
[u
] == deg
[v
] && u
> v
)swap(u
, v
);ve
[u
].push_back(i
);}for (int i
= 1; i
<= Max
; i
++) {for (int j
: ve
[i
])vis
[sB
[j
]]= sC
[j
];for (int j
: ve
[i
])for (int k
: ve
[sB
[j
]])if (vis
[sB
[k
]]) {ta
= vis
[sB
[k
]];tb
= sC
[j
];tc
= sC
[k
];get(i
, sB
[j
], sB
[k
]);}for (int j
: ve
[i
])vis
[sB
[j
]]= 0;}
}
void getans(int x
)
{for (int l
= 1, r
; l
<= x
; l
= r
+ 1) {r
= (x
/ (x
/ l
));p
[x
]= (p
[x
] + 1ll * (r
- l
+ 1) * (x
/ l
) % mod
) % mod
;}
}
int main()
{scanf("%d", &T
);get_mu();for (int i
= 1; i
<= 1e5; i
++)getans(i
);while (T
--) {scanf("%d%d%d", &A
, &B
, &C
);ans
= cnt
= 0;Max
= max(max(A
, B
), C
);for (int i
= Max
; i
>= 1; i
--) {for (int j
= i
; j
<= Max
; j
+= i
) {if (!mu
[j
])continue;for (int k
= j
; 1ll*k
*j
/i
<= 1ll * Max
; k
+= i
) {if (!mu
[k
])continue;if (!mp
[j
][k
]) {
++cnt
;sA
[cnt
]= j
;sB
[cnt
]= k
;sC
[cnt
]= 1ll*j
*k
/ i
;deg
[j
]++;deg
[k
]++;mp
[j
][k
]= 1;}}}}solve();for (int i
= 1; i
<= Max
; i
++) {ve
[i
].clear();mp
[i
].clear();deg
[i
]= 0;}printf("%d\n", (ans
+ mod
) % mod
);}return 0;
}
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