結果
問題 | No.1043 直列大学 |
ユーザー |
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提出日時 | 2020-05-01 22:36:34 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 264 ms / 2,000 ms |
コード長 | 2,463 bytes |
コンパイル時間 | 1,226 ms |
コンパイル使用メモリ | 99,980 KB |
実行使用メモリ | 83,680 KB |
最終ジャッジ日時 | 2024-12-22 19:35:30 |
合計ジャッジ時間 | 5,997 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 28 |
ソースコード
#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <deque>#include <list>#include <queue>#include <stack>#include <vector>#include <utility>#include <algorithm>#include <map>#include <set>#include <complex>#include <cmath>#include <limits>#include <climits>#include <ctime>#include <cassert>#include <numeric>#include <functional>#include <bitset>using namespace std;using lint = long long int;long long int INF = 1001001001001001LL;int inf = 1000000007;long long int MOD = 1000000007LL;double PI = 3.1415926535897932;template<typename T1,typename T2>inline void chmin(T1 &a,const T2 &b){if(a>b) a=b;}template<typename T1,typename T2>inline void chmax(T1 &a,const T2 &b){if(a<b) a=b;}#define ALL(a) a.begin(),a.end()#define RALL(a) a.rbegin(),a.rend()/* do your best */vector<lint> culc(vector<lint> a) {int n = a.size();lint m = 100 * 1000;vector<vector<lint>> dp(n + 1, vector<lint> (m + 1, 0));dp[0][0] = 1;for (int i = 0; i < n; i++) {for (int j = 0; j <= m; j++) {if (dp[i][j] == 0) continue;dp[i + 1][j] += dp[i][j];dp[i + 1][j] %= MOD;dp[i + 1][j + a[i]] += dp[i][j];dp[i + 1][j + a[i]] %= MOD;}}return dp[n];}// 抽象累積和// 構築 O(n), get O(1)template<typename T>struct CumSum{private:size_t n;vector<T> dat;public:CumSum(const vector<T> &v){n = v.size();dat.resize(n + 1, 0);for(size_t i = 0; i < n; i++){dat[i + 1] = dat[i] + v[i];dat[i + 1] %= MOD;}}T get(size_t r) const { // 0-indexed, [0. r)return dat[r];}T get(size_t l, size_t r){ // 0-indexed, [l, r)return (dat[r] - dat[l] + MOD) % MOD;}};int main() {int n, m; cin >> n >> m;vector<lint> v(n);vector<lint> r(m);for (int i = 0; i < n; i++) {cin >> v[i];}for (int i = 0; i < m; i++) {cin >> r[i];}lint a, b; cin >> a >> b;auto dp1 = culc(v);auto dp2 = culc(r);for (int i = 1; i <= 100 * 1000; i++) {dp1[i] += dp1[i - 1];dp1[i] %= MOD;}// 抵抗を決め打ちlint ans = 0;for (int r = 1; r <= 100 * 1000; r++) {// 範囲が決まlint lb = a * r;lint rb = b * r;if (100 * 1000 < lb) {continue;}rb = min(rb, 100 * 1000LL);lint sum = (dp1[rb] - dp1[lb - 1] + MOD) % MOD;ans += sum * dp2[r];ans %= MOD;}cout << ans << endl;return 0;}