#include<cstdio> #include<iostream> #include<algorithm> #include<string> #include<vector> #include<unordered_set> using namespace std; typedef vector<int> VI; typedef long long LL; #define FOR(x, b, e) for(int x=b; x<=(e); ++x) #define FORD(x, b, e) for(int x=b; x>=(e); --x) #define REP(x, n) for(int x=0; x<(n); ++x) #define VAR(v, n) __typeof(n) v = (n) #define ALL(c) (c).begin(), (c).end() #define SIZE(x) ((int)(x).size()) #define FOREACH(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i) #define PB push_back #define ST first #define ND second int main() { ios_base::sync_with_stdio(0); int n, k, q; cin >> n >> k >> q; VI a(n); REP(i, n) cin >> a[i]; vector<LL> ps(n + 1); vector<LL> dp(n + 1); REP(i, n) ps[i + 1] += ps[i] + a[i]; REP(i, q) { int l, r; cin >> l >> r; LL sum = ps[l - 1 + k] - ps[l - 1]; dp[l + k] = max(sum, 0ll); for(int i = k + 1; i < r - l + 2; i++) { sum += a[l + i - 2]; sum -= a[l + i - k - 2]; dp[l + i] = max(sum + (l + i - k < l + k ? 0 : dp[l + i - k]), dp[l + i - 1]); } cout << dp[r + 1] << '\n'; } return 0; }
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 | #include<cstdio> #include<iostream> #include<algorithm> #include<string> #include<vector> #include<unordered_set> using namespace std; typedef vector<int> VI; typedef long long LL; #define FOR(x, b, e) for(int x=b; x<=(e); ++x) #define FORD(x, b, e) for(int x=b; x>=(e); --x) #define REP(x, n) for(int x=0; x<(n); ++x) #define VAR(v, n) __typeof(n) v = (n) #define ALL(c) (c).begin(), (c).end() #define SIZE(x) ((int)(x).size()) #define FOREACH(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i) #define PB push_back #define ST first #define ND second int main() { ios_base::sync_with_stdio(0); int n, k, q; cin >> n >> k >> q; VI a(n); REP(i, n) cin >> a[i]; vector<LL> ps(n + 1); vector<LL> dp(n + 1); REP(i, n) ps[i + 1] += ps[i] + a[i]; REP(i, q) { int l, r; cin >> l >> r; LL sum = ps[l - 1 + k] - ps[l - 1]; dp[l + k] = max(sum, 0ll); for(int i = k + 1; i < r - l + 2; i++) { sum += a[l + i - 2]; sum -= a[l + i - k - 2]; dp[l + i] = max(sum + (l + i - k < l + k ? 0 : dp[l + i - k]), dp[l + i - 1]); } cout << dp[r + 1] << '\n'; } return 0; } |