这很显然是一道dp题。
令dp[i][j]表示第 i 分钟末，疲劳度为 j 是的最大跑步距离，则
dp[i][0] = max(dp[i - 1][0], max(dp[i - j][j]))
dp[i][j] = max(dp[i - 1][j - 1] + a[i])
因为题中说即使疲劳值为0了，仍可以休息，所以dp[i][0]也可以从dp[i - 1][0]转换过来。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef double db; 17 const int INF = 0x3f3f3f3f; 18 const int eps = 1e-8; 19 const int maxn = 1e4 + 5; 20 inline ll read() 21 { 22 ll ans = 0; 23 char ch = getchar(), last = ' '; 24 while(!isdigit(ch)) {last = ch; ch = getchar();} 25 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 26 if(last == '-') ans = -ans; 27 return ans; 28 } 29 inline void write(ll x) 30 { 31 if(x < 0) x = -x, putchar('-'); 32 if(x >= 10) write(x / 10); 33 putchar(x % 10 + '0'); 34 } 35 36 int n, m, a[maxn]; 37 int dp[maxn][maxn]; 38 39 int main() 40 { 41 n = read(); m = read(); 42 for(int i = 1; i <= n; ++i) a[i] = read(); 43 for(int i = 1; i <= n; ++i) 44 { 45 dp[i][0] = dp[i - 1][0]; 46 for(int j = 1; j <= min(i, m); ++j) dp[i][0] = max(dp[i][0], dp[i - j][j]); 47 for(int j = 1; j <= min(i, m); ++j) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + a[i]); 48 } 49 write(dp[n][0]); enter; 50 return 0; 51 }