結果
| 問題 | No.752 mod数列 |
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2026-01-23 23:30:40 |
| 言語 | C++17 (gcc 15.2.0 + boost 1.89.0) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 2,929 bytes |
| 記録 | |
| コンパイル時間 | 1,175 ms |
| コンパイル使用メモリ | 100,156 KB |
| 実行使用メモリ | 16,200 KB |
| 最終ジャッジ日時 | 2026-01-23 23:30:47 |
| 合計ジャッジ時間 | 7,436 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
(要ログイン)
| ファイルパターン | 結果 |
|---|---|
| sample | -- * 3 |
| other | AC * 17 TLE * 1 -- * 13 |
ソースコード
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 100010;
struct Block {
int l, r, frst_ele, c;
};
int p, q;
PII pp[N];
vector<Block> blocks;
vector<int> ls, rs;
LL Calc1(int id, int l, int r) {
int ele = blocks[id].frst_ele, c = blocks[id].c, cl = blocks[id].l;
int l_val = ele - c * (l - cl);
int r_val = ele - c * (r - cl);
return 1LL * (l_val + r_val) * (r - l + 1) / 2;
}
LL Calc2(int id, int pos, int op) {
int ele = blocks[id].frst_ele, c = blocks[id].c, cr = blocks[id].r, cl = blocks[id].l;
if (op == 0) { // 左块
int l_val = ele - c * (pos - cl);
int r_val = ele - c * (cr - cl);
if (c == 0) return 1LL * l_val * (cr - pos + 1);
return 1LL * (l_val + r_val) * (cr - pos + 1) / 2;
} else {
int l_val = ele;
int r_val = ele - c * (pos - cl);
if (c == 0) return 1LL * l_val * (pos - cl + 1);
return 1LL * (l_val + r_val) * (pos - cl + 1) / 2;
}
}
LL Calc3(int lb, int rb) {
LL ret = 0LL;
for (int i = lb; i <= rb; ++i) {
int ele = blocks[i].frst_ele, c = blocks[i].c, cl = blocks[i].l, cr = blocks[i].r;
int l_val = ele;
int r_val = ele - c * (cr - cl);
if (c == 0) ret += 1LL * l_val * (cr - cl + 1);
else ret += 1LL * (l_val + r_val) * (cr - cl + 1) / 2;
}
return ret;
}
int main() {
// freopen("mod.in", "r", stdin);
// freopen("mod.out", "w", stdout);
scanf("%d%d", &p, &q);
for (int i = 1; i <= q; ++i) scanf("%d%d", &pp[i].first, &pp[i].second);
int cl = 1, cr = 1;
for (int i = 0; cr < p; ++i) {
if (i != 0) {
cl = cr + 1;
cr = p / (p / cl);
}
if (cl != cr) {
blocks.push_back({ cl, cr, p % cl, (p % cl) - (p % (cl + 1)) });
} else {
blocks.push_back({ cl, cr, p % cl, 0 });
}
ls.push_back(cl), rs.push_back(cr);
}
for (int i = 1; i <= q; ++i) {
LL ans = 0LL;
if (pp[i].second > p) {
ans += 1LL * (pp[i].second - max(pp[i].first, p + 1) + 1) * p;
pp[i].second = min(pp[i].second, p);
}
if (pp[i].first <= p) {
int lb = upper_bound(ls.begin(), ls.end(), pp[i].first) - ls.begin() - 1;
int rb = lower_bound(rs.begin(), rs.end(), pp[i].second) - rs.begin();
if (lb == rb) {
ans += Calc1(lb, pp[i].first, pp[i].second); // 一个块内
} else {
ans += Calc2(lb, pp[i].first, 0); // 最左块(部分)
if (rb >= lb + 2) {
ans += Calc3(lb + 1, rb - 1); // 中间块(完整)
}
ans += Calc2(rb, pp[i].second, 1); // 最右块(部分)
}
}
printf("%lld\n", ans);
}
return 0;
}