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Codeforces round #268 (Div. 2) Question Link A: You just need to mark it out. B: Just judge the enumeration time one by one. C: Create a number. Manually construct the values 4 and 5 respectively, and then subtract 1 from the number 2 at a time, and then multiply the original number without changing, 4. The following cannot be constructed. D: greedy. Sort them fi

) { -memset (Vis,false,sizeof(Vis)); $Memset (DP,0,sizeof(DP)); $ -res =0; DFS (1); dp[1] =Res; -memset (Vis,false,sizeof(Vis)); theDFS2 (1); - Wuyi intMN = INF, p =0; the for(intI=1; ii) { - if(MN >=Dp[i]) { WuMN = Dp[i]; p =i; - } About } $printf ("%d\n", MN); - for(intI=1; ii) { - if(i = =p) { -printf ("%d\n", i); Break; A } + if(Dp[i] = = MN) printf ("%d", i); the } - } $ the intMainvoid) {//codeforces Round #135 (

Codeforces Round #262 (Div. 2 ),Codeforces Round #262 (Div. 2) A: You can directly simulate the problem. B: Since the size of s (x) is as large as 1e9, the number and the maximum value are 81. In this way, if we enumerate s (x), we only need to enumerate 1 to 81, then, after calculating the x value, determine whether t

Codeforces Round #301 (Div. 2 ), Question transfer: Codeforces Round #301 (Div. 2) A. Combination Lock Water question, find the minimum number of moves, simply greedy AC code: #include B. School Marks It's also relatively simple, it's a little complicated. Let's look at the code. AC code: # Include

A long time did not do the game, the original dish now more vegetables, 1004 of the topic did not expect Div.2 after the game only one person to do, do the problem when the idea of the practice but time not much of their own flower code is not enough in the game time debug out, after the game to fill a pitch:Title Submit Address in this: http://acm.hdu.edu.cn/showproblem.php?pid=5565Const intINF =100000000

Topic Transmission: Codeforces Round #301 (Div. 2)A. Combination LockWater problem, to find the minimum number of moves, simple greedy canAC Code:#include B. School MarksAlso relatively simple, is a bit cumbersome, specific look at the code barAC Code:#include C. Ice CaveTest instructions: Very simple, is a n*m ice, some broken, go once will be trapped, some intact, but walk once broken, next time will be t

Topic Portal1 /*2 Test Instructions: Arranges the first string arbitrarily so that there is no more than A/b string appearing3 string processing/greed: Brute force finds the maximum non-overwriting a string and then dynamically obtains the optimal solution in the B string4 I 'm sick of it, I originally wanted to output up to a, then the most B, but it is not guaranteed to be the most: (5 */6#include 7#include 8#include string>9#include Ten#include On

Topic Portal1 /*2 Test Instructions: Give a series of name changes, ask what the last name became3 string Processing: Each time the input to the previous name of the printing, if not, it is the initial, Pos[m] array record the initial position4 the initial POS is added to each update, so the initial name is guaranteed to be updated, which is the only place to think about:)5 */6#include 7#include 8#include 9#include Ten#include One#include string> A#i

Topic Portal1 /*2 water Problem: Read the topic can do3 */4#include 5#include 6#include 7#include 8#include 9#include Ten#include string> One#include A#include -#include Set> - using namespacestd; the - Const intMAXN = 1e2 +Ten; - Const intINF =0x3f3f3f3f; - intA[MAXN][MAXN]; + - intMainvoid)//codeforces Round #308 (Div. 2) A. Vanya and Table + { A //freo

Topic Portal1 /*2 water problem: To find the total number of numbers, open long long unexpectedly WA several times, also did not change what again:)3 */4#include 5#include 6#include 7#include 8#include 9#include Ten#include string> One#include A#include -#include Set> - using namespacestd; the -typedefLong Longll; - Const intMAXN = 1e2 +Ten; - Const intINF =0x3f3f3f3f; + - intMainvoid)//codeforces Round #308 (