dp，參考了http://www.cppblog.com/hanfei19910905/archive/2012/06/30/180831.html；

定義兩組狀態 L1[i] 表示在 0-i 內的最大得分，L2[i] 表示從 i 出發回到 i 的最大得分，不難得到狀態轉移方程，最后枚舉 i 組合求出最大值；

# include <cstdio> # include <algorithm># define N 100005using namespace std;typedef long long int LL;int n, num[N]; LL L1[N], L2[N], R1[N], R2[N], ans;void init(void) {int i;scanf("%d", &n);for (i = 1; i < n; ++i)scanf("%d", &num[i]); }void solve(void) {int i;L1[0] = L2[0] = 0;for (i = 1; i < n; ++i){L2[i] = (num[i]>1 ? L2[i-1]+num[i]/2*2:0);if (num[i] & 0x1) L1[i] = L1[i-1] + num[i];else L1[i] = max(L2[i], L1[i-1]+num[i]-1);}R2[n-1] = R1[n-1] = 0;for (i = n-2; i >= 0; --i){R2[i] = (num[i+1]>1 ? R2[i+1]+num[i+1]/2*2:0);if (num[i+1] & 0x1) R1[i] = R1[i+1] + num[i+1];else R1[i] = max(R2[i], R1[i+1]+num[i+1]-1);}ans = 0;for (i = 0; i < n; ++i){ans = max(ans, max(max(L1[i], R1[i]), max(L1[i]+R2[i], L2[i]+R1[i])));}printf("%I64d\n", ans); }int main() {//freopen("in.txt", "r", stdin); init();solve();return 0; }

轉載于:https://www.cnblogs.com/JMDWQ/archive/2012/07/13/2590845.html

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