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  • [HDU

    [HDU - 5116 ] Everlasting L (计数DP,容斥)

    题目链接:HDU - 5116

    题目链接:

    题意:

    给定一个大小为(mathit n)的点集(mathit S),现在让求出有多少个集合对((A,B)),满足:

    img

    一个集合被称为Good,当且仅当满足:

    点集(mathit P) 中存在一个点((x,y)),满足:

    P = {(x, y), (x + 1, y), . . . , (x + a, y), (x, y + 1), . . . , (x, y + b)}(a, b ≥ 1)

    并且 gcd(a, b) = 1.

    思路:

    考虑容斥:

    设Good集合个数=a。

    设有交集Good集合对数=b。

    则答案(ans=a^2-b)

    问题转化为求(a,b)

    先利用DP求出数组:

    (f[i][j])代表(mathit i)([1,j])中多少个数互质。

    (g[i][j])代表([1,i])([1,j])中多少对数互质。

    (right[i][j])代表集合中点((i,j))右边有多少个连续的点。

    (up[i][j])代表集合中点((i,j))上边有多少个连续的点。

    那么我们对每一个点集中纯在的点((i,j)),枚举(kin[1,up[i][j]]),该点对(mathit a)的贡献即为(sum f[k][right[i][j]])

    利用同样的方式可以获得出dp数组(cnt[i][j]) 代表 点((i,j))在多少个Good集合的上升部分。

    数组(num[i][j]) 代表 点((i,j))在多少个Good集合的右拓部分。

    那么每一个点((i,j))(mathit b)的贡献即为(2 imes num[i][j]*cnt[i][j]-g[up[i][j][right[i][j]]]^2)

    代码:

    #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), '', 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("
    ");}}}
    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("
    ");}}}
    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 ? '
    ' : ' ');}}
    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 ? '
    ' : ' ');}}
    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
    ", icase, ans );
        }
        return 0;
    }
    
    
    
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  • 原文地址:https://www.cnblogs.com/qieqiemin/p/13864225.html
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