[HDU - 5116 ] Everlasting L （計數DP，容斥）
題目鏈接：HDU - 5116
題目鏈接：
題意：
給定一個大小爲\(\mathit n\)的點集\(\mathit S\)，現在讓求出有多少個集合對\((A,B)\)，滿足：
一個集合被稱爲Good，當且僅當滿足：
點集\(\mathit P\) 中存在一個點\((x,y)\),滿足：
P = {(x, y), (x + 1, y), . . . , (x + a, y), (x, y + 1), . . . , (x, y + b)}(a, b ≥ 1)
並且 gcd(a, b) = 1.
思路：
考慮容斥：
設Good集合個數=a。
設有交集Good集合對數=b。
則答案\(ans=a^2-b\)。
問題轉化爲求\(a,b\)。
先利用DP求出數組：
\(f[i][j]\)代表\(\mathit i\)與\([1,j]\)中多少個數互質。
\(g[i][j]\)代表\([1,i]\)與\([1,j]\)中多少對數互質。
\(right[i][j]\)代表集合中點\((i,j)\)右邊有多少個連續的點。
\(up[i][j]\)代表集合中點\((i,j)\)上邊有多少個連續的點。
那麼我們對每一個點集中純在的點\((i,j)\)，枚舉\(k\in[1,up[i][j]]\)，該點對\(\mathit a\)的貢獻即爲\(\sum f[k][right[i][j]]\)。
利用同樣的方式可以獲得出dp數組\(cnt[i][j]\) 代表 點\((i,j)\)在多少個Good集合的上升部分。
數組\(num[i][j]\) 代表 點\((i,j)\)在多少個Good集合的右拓部分。
那麼每一個點\((i,j)\)對\(\mathit b\)的貢獻即爲\(2\times num[i][j]*cnt[i][j]-g[up[i][j][right[i][j]]]^2\)。
代碼：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n;
ll ans;
ll vis[205][205];
ll _right[205][205];
ll up[205][205];
ll num[205][205];
ll cnt[205][205];
int len = 200;
ll f[205][205];
ll g[205][205];
int main()
{
#if DEBUG_Switch
freopen("C:\\code\\input.txt", "r", stdin);
#endif
//freopen("C:\\code\\output.txt","w",stdout);
int t;
t = readint();
repd(i, 1, len) {
repd(j, 1, len) {
f[i][j] = f[i][j - 1] + (gcd(i, j) == 1);
g[i][j] = g[i - 1][j] + f[i][j];
}
}
for (int icase = 1; icase <= t; ++icase) {
MS0(vis);
MS0(_right);
MS0(up);
MS0(cnt);
MS0(num);
n = readint();
repd(i, 1, n) {
int x = readint(), y = readint();
vis[x][y] = 1;
}
for (int i = len; i >= 1; --i) {
for (int j = len; j >= 1; --j) {
if (vis[i + 1][j] > 0) {
_right[i][j] = _right[i + 1][j] + 1;
}
if (vis[i][j + 1] > 0) {
up[i][j] = up[i][j + 1] + 1;
}
}
}
ll a = 0ll;
repd(i, 1, len) {
repd(j, 1, len) {
if (vis[i][j] == 0) {
continue;
}
ll temp = 0ll;
for (int k = up[i][j]; k >= 0; --k) {
temp += f[k][_right[i][j]];
cnt[i][j + k] += temp;
}
a += temp;
}
}
repd(i, 1, len) {
repd(j, 1, len) {
if (vis[i][j] == 0) {
continue;
}
ll temp = 0ll;
for (int k = _right[i][j]; k >= 0; --k) {
temp += f[k][up[i][j]];
num[i + k][j] += temp;
}
}
}
ll b = 0ll;
repd(i, 1, len) {
repd(j, 1, len) {
if (vis[i][j] > 0) {
b += num[i][j] * cnt[i][j] * 2;
b -= g[up[i][j]][_right[i][j]] * g[up[i][j]][_right[i][j]];
}
}
}
ll ans = a * a - b;
printf("Case #%d: %lld\n", icase, ans );
}
return 0;
}