#13：人十我一Orz——6

水题放送，写得依旧丑：

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 using namespace std;

 const int INF = 2e9 + ;
 typedef long long ll;
 typedef pair<int, int> P;
 int H, G;
 ll f[][][];
 int dishg[][];
 int dish[], disg[];
 vector<P> hh, gg;

 int sqr(int x) {
     return x * x;
 }

 int dis(vector<P> v1, int i, vector<P> v2, int j) {
     return sqr(v1[i].first - v2[j].first) + sqr(v1[i].second - v2[j].second);
 }

 void init() {
     for (int i = ; i <= H; i++)
         for (int j = ; j <= G; j++)
             dishg[i][j] = dis(hh, i, gg, j);
     f[][][] = INF;
     for (int i = ; i <= H; i++) {
         dish[i] = dis(hh, i-, hh, i);
         f[i][][] = f[i-][][] + dish[i];
         f[i][][] = INF;
     }
     for (int j = ; j <= G; j++) {
         disg[j] = dis(gg, j-, gg, j);
         f[][j][] = INF;
         f[][j][] = INF;
     }
 }

 ll dp() {
     for (int i = ; i <= H; i++) {
         for (int j = ; j <= G; j++) {
             ll a = min(f[i-][j][] + dish[i], f[i-][j][] + dishg[i][j]);
             ll b = min(f[i][j-][] + dishg[i][j], f[i][j-][] + disg[j]);
             f[i][j][] = a;
             f[i][j][] = b;
         }
     }
     return f[H][G][];
 }

 int main() {
     scanf("%d%d", &H, &G);
     hh.push_back(P(, ));
     gg.push_back(P(, ));
     for (int i = ; i <= H; i++) {
         int x, y;
         scanf("%d%d", &x, &y);
         hh.push_back(P(x, y));
     }
     for (int i = ; i <= G; i++) {
         int x, y;
         scanf("%d%d", &x, &y);
         gg.push_back(P(x, y));
     }
     init();
     printf("%lld\n", dp());
     return ;
 }

BZOJ4745

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 using namespace std;

 const int INF = 2e9 + ;
 typedef long long ll;
 typedef pair<int, int> P;
 int N, H, Delta;
 int f[][], eat[][];
 int pos1[];

 int dp() {
     for (int i = ; i <= H; i++) {
         int maxx = ;
         for (int j = ; j <= N; j++) {
             f[i][j] = f[i-][j];
             if (i > Delta) {
                 int t = i - Delta;
                 f[i][j] = max(f[i][j], f[t][pos1[t]]);
             }
             f[i][j] += eat[j][i];
             if (f[i][j] > maxx) {
                 maxx = f[i][j];
                 pos1[i] = j;
             }
         }
     }
     int ans = ;
     for (int i = ; i <= N; i++)
         ans = max(ans, f[H][i]);
     return ans;
 }

 int main() {
     scanf("%d%d%d", &N, &H, &Delta);
     for (int i = ; i <= N; i++) {
         int ni, pos;
         for (scanf("%d", &ni); ni; ni--) {
             scanf("%d", &pos);
             eat[i][pos]++;
         }
     }
     printf("%d\n", dp());
     return ;
 }

BZOJ1270

 #include <bits/stdc++.h>
 #define ri readint()
 #define gc getchar()
 #define ls p << 1
 #define rs p << 1 | 1
 using namespace std;

 typedef long long ll;
 const int maxn = ;
 int n, m;
 struct Seg {
     int l, r;
     bool fixed;
     ll sum;
 }t[maxn << ];

 inline int readint() {
     int x = , s = , c = gc;
     while (c <= )    c = gc;
     if (c == '-')    s = -, c = gc;
     for (; isdigit(c); c = gc)
         x = x *  + c - ;
     return x * s;
 }

 void build(int l, int r, int p) {
     t[p].l = l, t[p].r = r;
     if (l == r) {
         t[p].sum = ri;
         t[p].fixed = t[p].sum <= 1ll;
         return;
     }
     int mid = (l + r) >> ;
     build(l, mid, ls);
     build(mid+, r, rs);
     t[p].sum = t[ls].sum + t[rs].sum;
     t[p].fixed = t[ls].fixed && t[rs].fixed;
 }

 void Update(int l, int r, int p) {
     if (t[p].fixed)    return;
     if (t[p].l == t[p].r) {
         t[p].sum = sqrt(t[p].sum);
         t[p].fixed = t[p].sum <= 1ll;
         return;
     }
     int mid = (t[p].l + t[p].r) >> ;
     if (l <= mid)    Update(l, r, ls);
     if (mid < r)    Update(l ,r, rs);
     t[p].sum = t[ls].sum + t[rs].sum;
     t[p].fixed = t[ls].fixed && t[rs].fixed;
 }

 ll Query(int l, int r, int p) {
     if (l <= t[p].l && t[p].r <= r)
         return t[p].sum;
     int mid = (t[p].l + t[p].r) >> ;
     ll res = 0ll;
     if (l <= mid)    res += Query(l, r, ls);
     if (mid < r)    res += Query(l, r, rs);
     return res;
 }

 int main() {
     n = ri;
     build(, n, );
     for (m = ri; m; m--) {
         int x = ri, l = ri, r = ri;
         if (x == ) {
             printf("%lld\n", Query(l, r, ));
         } else {
             Update(l, r, );
         }
     }
     return ;
 }

BZOJ3211

cf818E，计数不一定非要乘法原理，枚举标杆累加。此题性质符合尺取，l和r可不断后移。

 #include <bits/stdc++.h>
 using namespace std;

 const int maxn = 1e5 + ;
 int n, k, a[maxn];
 int pri[], cnt[], t;
 int b[maxn][];

 void div(int k) {
     for (int i = ; i * i <= k; i++) {
         if (k % i == ) {
             pri[++t] = i;
             while (k % i == ) {
                 k /= i;
                 cnt[t]++;
             }
         }
     }
     if (k > ) {
         pri[++t] = k;
         cnt[t] = ;
     }
 }

 long long two_point() {
     int l = , r = ;
     long long ans = ;
     while (l <= n) {
         bool valid = false;
         while (r < n && !valid) {
             r++;
             valid = true;
             for (int j = ; j <= t; j++)
                 valid &= (b[r][j] - b[l-][j]) >= cnt[j];
         }
         if (!valid)    break;
         while (l <= r && valid) {
             ans += n - r + ;
             l++;
             for (int j = ; j <= t; j++)
                 valid &= (b[r][j] - b[l-][j]) >= cnt[j];
         }
     }
     return ans;
 }

 int main() {
     scanf("%d%d", &n, &k);
     div(k);
     for (int i = ; i <= n; i++) {
         scanf("%d", &a[i]);
         for (int j = ; j <= t; j++) {
             while (a[i] % pri[j] == ) {
                 a[i] /= pri[j];
                 b[i][j]++;
             }
             b[i][j] += b[i-][j];
         }
     }
     cout << two_point() << endl;
     return ;
 }

cf818F，我强猜一下结论的原因：难道是因为完全图边多、链桥多，所以边多桥多就凑一起？

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long ll;
 int q;
 ll n;

 ll judge(ll k) {
     return n - k + min(n - k, (k*k - k) / );
 }

 int main() {
     ios_base::sync_with_stdio(false);
     cin.tie();
     for (cin >> q; q; q--) {
         cin >> n;
         ll l = , r = n;
         while (l < r-) {
             ll lmid = (l + r) >> ;
             ll rmid = (lmid + r) >> ;
             if (judge(lmid) < judge(rmid))
                 l = lmid;
             else    r = rmid;
         }
         cout << max(judge(r), judge((l + r) >> )) << "\n";
     }
     return ;
 }

cf822D，遍历素因子的技巧值得学习。以及貌似第一发猜的结论好像是对的但是我写挂了……but还是正解优雅。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long ll;
 const int maxn = 5e6 + ;
 const int mod = 1e9 + ;
 const ll INF = 1e18;
 int t, l, r;
 ll f[maxn];
 int primes[maxn];

 void init() {
     for (int i = ; i <= r; i++) {
         primes[i] = i;
         f[i] = INF;
     }
     for (int i = ; i * i <= r; i++) {//不采集素数时不需遍历maxn
         if (primes[i] == i) {
             for (int j = i * i; j <= r; j += i)
                 primes[j] = min(primes[j], i);
         }
     }
     for (int i = ; i <= r; i++) {
         for (int j = i; j != ; j /= primes[j]) {//遍历素因子
             f[i] = min(f[i], f[i / primes[j]] + 1LL * i * (primes[j] - ) / 2LL);
         }
     }
 }

 int main() {
     scanf("%d%d%d", &t, &l, &r);
     init();
     ll ans = , tt = ;
     for (int i = l; i <= r; i++) {
         f[i] %= mod;
         ans = (ans + tt * f[i]) % mod;
         tt = tt * t % mod;
     }
     cout << ans << endl;
     return ;
 }