#include <bits/stdc++.h> #define FOR(i, b, e) for (int i = (b); i <= (e); i++) #define FORD(i, b, e) for (int i = (e); i >= (b); i--) #define MP make_pair #define FS first #define ND second #define PB push_back #define ALL(x) x.begin(), x.end() using namespace std; using LL = long long; using PII = pair<int,int>; using PLL = pair<LL,LL>; using VI = vector<int>; using VLL = vector<LL>; template<class T> using V = vector<T>; template<class T> int sz(const T& a) { return (int)a.size(); } template<class T> void amin(T& a, T b) { a = min(a, b); } template<class T> void amax(T& a, T b) { a = max(a, b); } const int inf = 1e9; const LL infl = 1e18; const int N = 5e3; const int mod = int(1e9) + 7; int n, ans, a[N + 1], dp[N + 2]; void add(int& x, int y) { x = (x + y) % mod; } int main() { scanf("%d", &n); FOR(i, 1, n) scanf("%d", &a[i]); sort(a + 1, a + n + 1); dp[0] = 1; FOR(i, 1, n) { FORD(j, a[i] - 1, N + 1) { add(dp[min(j + a[i], N + 1)], dp[j]); } } FOR(i, 1, N + 1) add(ans, dp[i]); printf("%d\n", ans); }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 | #include <bits/stdc++.h> #define FOR(i, b, e) for (int i = (b); i <= (e); i++) #define FORD(i, b, e) for (int i = (e); i >= (b); i--) #define MP make_pair #define FS first #define ND second #define PB push_back #define ALL(x) x.begin(), x.end() using namespace std; using LL = long long; using PII = pair<int,int>; using PLL = pair<LL,LL>; using VI = vector<int>; using VLL = vector<LL>; template<class T> using V = vector<T>; template<class T> int sz(const T& a) { return (int)a.size(); } template<class T> void amin(T& a, T b) { a = min(a, b); } template<class T> void amax(T& a, T b) { a = max(a, b); } const int inf = 1e9; const LL infl = 1e18; const int N = 5e3; const int mod = int(1e9) + 7; int n, ans, a[N + 1], dp[N + 2]; void add(int& x, int y) { x = (x + y) % mod; } int main() { scanf("%d", &n); FOR(i, 1, n) scanf("%d", &a[i]); sort(a + 1, a + n + 1); dp[0] = 1; FOR(i, 1, n) { FORD(j, a[i] - 1, N + 1) { add(dp[min(j + a[i], N + 1)], dp[j]); } } FOR(i, 1, N + 1) add(ans, dp[i]); printf("%d\n", ans); } |