The 15th Heilongjiang Provincial Collegiate Programming Contest
阿新 • • 發佈：2021-07-16
https://codeforces.com/gym/102803
A: August
猜結論的題目，nt
B：Bills of Paradise
線段樹 四種操作：
1、查詢區間第一個比x大的數
2、刪除區間第一個比x大的數
3、區間所有小於x的數求和
4、將區間所有原來小於x的數還原
顯然權值線段樹維護區間最大值可以實現 1和2功能
可以通過維護區間和與區間最小值最大值實現 3功能
通過陣列記錄建樹時整棵樹的狀態可以實現 4功能 通過lz標記可以優化
const int maxn = 1e6 + 10; unsigned long long k1, k2; unsigned long long xorShift128Plus() { unsigned long long k3 = k1, k4 = k2; k1 = k4; k3 ^= k3 << 23; k2 = k3 ^ k4 ^ (k3 >> 17) ^ (k4 >> 26); return k2 + k4; } int n, m; long long a[1000001], b[maxn], v[maxn]; char s[3]; struct seg{ int t[maxn << 2], sum[maxn << 2], tv[maxn << 2], ts[maxn << 2], mx[maxn << 2], mn[maxn << 2]; bool lz[maxn << 2]; #define root 1, 1, m #define node rt, l, r #define lr rt << 1 #define rr rt << 1 | 1 #define mid (l + r >> 1) #define lson lr, l, mid #define rson rr, mid + 1, r inline void push(int rt) { t[rt] = t[lr] + t[rr]; sum[rt] = sum[lr] + sum[rr]; mx[rt] = max(mx[lr], mx[rr]); mn[rt] = min(mn[lr], mn[rr]); } inline void upd(int rt, int l, int r) { lz[rt] = 1; t[rt] = tv[rt]; sum[rt] = ts[rt]; mx[rt] = b[r]; mn[rt] = b[l]; } inline void down(int rt, int l, int r) { lz[rt] = 0; upd(lson); upd(rson); } void build(int rt, int l, int r) { if(l == r) { tv[rt] = t[rt] = v[l]; ts[rt] = sum[rt] = b[l] * v[l]; mx[rt] = mn[rt] = b[l]; return ; } build(lson); build(rson); tv[rt] = tv[lr] + tv[rr]; ts[rt] = ts[lr] + ts[rr]; push(rt); } int queryup(int rt, int l, int r, int x) { if(l == r) return b[l]; if(lz[rt]) down(node); if(mx[lr] >= x) return queryup(lson, x); else if(mx[rr] >= x) return queryup(rson, x); return 1e12; } int querysum(int rt, int l, int r, int x) { if(mn[rt] > x) return 0; if(mx[rt] <= x) return sum[rt]; if(lz[rt]) down(node); return querysum(lson, x) + querysum(rson, x); } void reinit(int rt, int l, int r, int x) { //printf("%lld %lld %lld %lld %lld\n", rt, l, r, b[l], b[r]); if(b[l] > x) return ; if(b[r] <= x) { upd(node); return ; } if(lz[rt]) down(node); else { reinit(lson, x); reinit(rson, x); } push(rt); } void change(int rt, int l, int r, int x) { if(l == r) { t[rt] --; sum[rt] -= b[l]; if(t[rt] == 0) mx[rt] = 0, mn[rt] = 0x3f3f3f3f3f3f3f3f; return ; } if(lz[rt]) down(node); if(mx[lr] >= x) change(lson, x); else if(mx[rr] >= x) change(rson, x); push(rt); } }T; void gen() { scanf("%d %llu %llu", &n, &k1, &k2); for (int i = 1; i <= n; i++) { a[i] = xorShift128Plus() % 999999999999 + 1; } sort(a + 1, a + 1 + n); m = 1; int sum = 0; for(int i = 1; i <= n; ++ i) { v[m] ++; sum += a[i]; if(a[i] != a[i + 1]) b[m++] = a[i]; } m--; T.build(root); //for(int i = 1; i <= m; ++ i) printf("%lld%c", b[i], " \n"[i == m]); int q = rd(); //printf("%lld %lld\n", T.querysum(root, 1e13), sum); for(int i = 1; i <= q; ++ i) { scanf("%s", s); int x = rd(); if(s[0] == 'F') printf("%lld\n", T.queryup(root, x)); else if(s[0] == 'D') T.change(root, x); else if(s[0] == 'C') printf("%lld\n", T.querysum(root, x)); else T.reinit(root, x); //printf("%lld\n", T.querysum(root, 1e12)); } } signed main() { gen(); return 0; }
C： Cornelia Street
超級無敵大水題，看起來什麼迴圈節好像要kmp，複雜度都算不出來，結果暴力列舉N^2隨便過
const int maxn = 1e6 + 10; int n, m, len, pos; char s[maxn]; bool ok(int l, int &pos) { int nowpos = -1; pos = -1; for(int i = l; i < len; ++ i) if(s[i % l] != s[i]) { pos = i - i % l; break; } if(pos == -1 || pos + 2 * l > len) return 0; for(int i = pos + l; i < len; ++ i) if(s[pos + i % l] != s[i]) { nowpos = i - i % l; if(nowpos + l > len) return 0; for(int j = nowpos; j < len; ++ j) if(s[j % l] != s[j]) return 0; break; } if(nowpos == -1) return 0; return 1; } signed main() { scanf("%s", s); len = strlen(s); for(int i = 1, up = len / 3; i <= up; ++ i) { if(ok(i, pos)) { for(int j = 0; j < i; ++ j) printf("%c", s[j]); printf(" "); for(int j = 0; j < i; ++ j) printf("%c", s[j + pos]); puts(""); return 0; } } return 0; }
F：False God
nt dp
讀題以為是互吃，我的回合碰到你就能吃你
實際上是隻要小兵碰到你就會被吃，不管是你主動還是被動，那直接按位置排序，n^2dp暴力轉移狀態
struct node{ int x, y; bool operator < (const node &a) const { return y == a.y ? x < a.x : y > a.y; } }a[maxn]; int dp[maxn]; signed main() { int t = rd(); while(t--) { int sx = rd(), sy = rd(); int n = rd(); for(int i = 1; i <= n; ++ i) a[i].x = rd(), a[i].y = rd(); sort(a + 1, a + 1 + n); //puts(""); for(int i = 1; i <= n; ++ i) { dp[i] = 1; // printf("%lld %lld\n", a[i].x, a[i].y); for(int j = 1; j < i; ++ j) { if(abs(a[i].x - a[j].x) - 1 <= a[j].y - a[i].y) dp[i] = max(dp[i], dp[j] + 1); } } int ans = 0; //for(int i = 1; i <= n; ++ i) printf("%lld%c", dp[i], " \n"[i == n]); for(int i = 1; i <= n; ++ i) if(sy <= a[i].y + 1 && abs(sx - a[i].x) - 1 <= a[i].y - sy) ans = max(ans, dp[i]); printf("%lld\n", ans); } return 0; }
G: Goodbye
簡單思維題
H：Hate That You Know Me
給\(k\in（0,1,2,3)\)，求
\[\sum_{i = 1}^n \sum_{d|i}d^k \]顯然轉化為整除分塊，\(d^k\) 通過公式O(1)求
\(mod = 1<<64\)顯然是unsigned long long 的自然溢位
const int mod = (1 << 64);
//n * n * (n + 1) * (n + 1) / 4
inline int sum(int sta, int n) {
if(sta == 3) {
if(n & 1) return (n + 1) / 2 * ((n + 1) / 2) * n * n;
else return n / 2 * (n / 2) * (n + 1) * (n + 1);
}
else if(sta == 2) {
if(n & 1) {
if(n % 3 == 0) return (n + 1) / 2 * (n / 3) * (2 * n + 1);
else if((n + 1) % 3 == 0) return (n + 1) / 6 * n * (2 * n + 1);
else return (n + 1) / 2 * n * ((2 * n + 1) / 3);
}
else {
if(n % 3 == 0) return n / 6 * (n + 1) * (2 * n + 1);
else if((n + 1) % 3 == 0) return (n + 1) / 3 * (n / 2) * (2 * n + 1);
else return (2 * n + 1) / 3 * (n / 2) * (n + 1);
}
}
else if(sta) {
if(n & 1) return (n + 1) / 2 * n;
else return n / 2 * (n + 1);
}
else return n;
}/*
inline int sum(int sta, int n) {
if(sta == 3) return n * n * (n + 1) * (n + 1) / 4;
else if(sta == 2) return (2 * n + 1) * (n) * (n + 1) / 6;
else if(sta) return n * (n + 1) / 2;
else return n;
}*/
inline void qmod(int &a) {
if(a >= mod) a -= mod;
}
void solve3() {//n * (n + 1) * (2n + 1) / 6
int a = rd(), b = rd(), n = rd();
int ans1 = 0, ans2 = 0;
for(int l = 1, r; l <= n; l = r + 1) {
r = n / (n / l);
ans1 += n / l * (sum(a, r) - sum(a, l - 1));
ans2 += n / l * (sum(b, r) - sum(b, l - 1));
}
printf("%llu\n", ans1 ^ ans2);
}
signed main() {
solve3();
return 0;
}
K： Keeping A Secret
滾遠點謝謝
L：Let's Get Married
找規律構造