CodeChef December Challenge 2013 解题报告
2013-12-08 01:19
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Problem A Magic Pairs
水体一枚,其实就是求1-n-1的和。用long long 就行。
代码如下:
View Code
水体一枚,其实就是求1-n-1的和。用long long 就行。
代码如下:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <stack> #include <queue> #include <vector> #include <utility> #define LEN 100100 #define INF 0x7fffffff #define ll long long using namespace std; int n, k; ll num[LEN], pre[LEN], las[LEN]; void init(){ memset(pre, 0 ,sizeof pre); memset(las, 0 ,sizeof las); for(int i=1; i<=n; i++){ pre[i] = pre[i-1]<0?num[i]:pre[i-1]+num[i]; } pre[0] = -INF; for(int i=1; i<=n; i++)pre[i] = max(pre[i-1], pre[i]); for(int i=n; i>=1; i--){ las[i] = las[i+1]<0?num[i]:las[i+1]+num[i]; } las[n+1] = -INF; for(int i=n; i>=1; i--)las[i] = max(las[i+1], las[i]); } int main() { // freopen("in.txt", "r", stdin); int T; scanf("%d", &T); while(T--){ scanf("%d%d", &n, &k); for(int i=1; i<=n; i++){ cin >> num[i]; } init(); ll ans = -INF; for(int i=1; i+k<n; i++){ ans = max(ans, pre[i]+las[i+k+1]); } cout << ans << endl; } return 0; }
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