Algorithm/DP(동적 계획법)40 [Dovelet][DP] 01knapsack 01knapsackhttp://59.23.113.171/30stair/01knapsack/01knapsack.php?pname=01knapsack1234567891011121314151617181920212223242526272829303132333435363738#include int Dp[101][10010] = {}; struct jjj{ int weight; int value;}; int main(void){ int N; int jNum; jjj jew[110] = {}; scanf("%d", &N); scanf("%d", &jNum); for (int i = 0; i 2016. 4. 12. [백준][2240번][DP] 자두나무 자두나무https://www.acmicpc.net/problem/22401234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253#include int max(int a, int b){ return a > b ? a : b;} int main(void){ int T, W; int Dp[2][1010][33] = {}; int jado[1010] = {}; int maxResult = 0; scanf("%d %d", &T, &W); for (int i = 1; i 2016. 4. 11. [백준][9095번][DP] 1,2,3 더하기 1,2,3 더하기https://www.acmicpc.net/problem/90951234567891011121314151617181920212223242526#include int main(void){ int testcase; scanf("%d", &testcase); while (testcase--){ int Dp[15] = {}; Dp[0] = 1; int K; scanf("%d", &K); for (int i = 1; i =0) Dp[i] += Dp[i - 1]; if (i - 2 >= 0) Dp[i] += Dp[i - 2]; if (i - 3 >= 0) Dp[i] += Dp[i - 3]; } printf("%d\n", Dp[K]); } return 0;}cs 1. 이 문제 같은경우는 1,2,3 만.. 2016. 4. 10. [백준][2294번][DP] 동전 2 동전 2https://www.acmicpc.net/problem/229412345678910111213141516171819202122232425262728293031323334353637383940414243444546474849#include int main(void){ int n, k; int Dp[10001] = {}; int coin[101]; scanf("%d %d", &n, &k); for (int i = 0; i 2016. 4. 9. [백준][2293번][DP] 동전 1 동전 1https://www.acmicpc.net/problem/229312345678910111213141516171819202122232425262728293031323334353637#include int Dp[2][10001] = {}; int main(void){ int n, k; int coin[101]; scanf("%d %d", &n, &k); for (int i = 0; i 2016. 4. 9. [ALGOSPOT][DP] PI PI https://www.algospot.com/judge/problem/read/PI1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192939495969798#include int a[10100]; int three(int num){ int nando = 10; if (a[num - 2] == a[num - 1] && a[num - 1] == a[num]) nando = 1; else if (a[num - 2] + 2 == a[num - 1] .. 2016. 4. 6. 이전 1 ··· 3 4 5 6 7 다음 728x90
