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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1941 Accepted Submission(s): 938
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #define min(a, b) a<b?a:b 5 using namespace std; 6 int main() 7 { 8 int t; 9 int num1[2020], num2[2020], dp[2020]; 10 scanf("%d", &t); 11 while(t--) 12 { 13 int m; 14 scanf("%d", &m); 15 for(int i = 1; i <= m; i++) 16 scanf("%d", &num1[i]); 17 for(int i = 1; i <= m-1; i++) 18 scanf("%d", &num2[i]); 19 memset(dp, 0, sizeof(dp)); 20 dp[1] = num1[1]; 21 for(int i = 2; i <= m; i++) 22 dp[i] = min(dp[i-1] + num1[i], dp[i-2] + num2[i-1]); 23 int b, c, d; 24 b = dp[m] % 60; 25 c = dp[m] / 60 % 60; 26 d = dp[m] /3600; 27 d += 8; 28 char c1, c2; 29 if(d < 12) 30 { 31 c1 = ‘a‘; 32 c2 = ‘m‘; 33 } 34 else 35 { 36 d %= 12; 37 c1 = ‘p‘; 38 c2 = ‘m‘; 39 } 40 printf("%02d:%02d:%02d %c%c\n", d, c, b, c1, c2); 41 } 42 return 0; 43 }
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原文地址:http://www.cnblogs.com/fengshun/p/4737650.html