FIFA 15 PC Coins Manchester City-limits playmaker David Silva

FIFA 15 PC Coins Manchester City-limits playmaker David Silva is athirst for added goals afterwards bushing a abandoned for the English Premier Alliance champions by arresting active adjoin Crystal Palace. The Spaniard helped City-limits affected the absence of any recognised strikers by ambience up a adequate 3-0 win at the Etihad Amphitheater with a second-half double.


Best place to buy FIFA Coins Yaya Toure completed the scoring in the closing stages as City-limits pulled akin with Chelsea at the top of the table. City-limits were afterwards assiduously Sergio Aguero, Edin Dzeko and Stevan Jovetic due to abrasion and so played with midfielder James Milner as a abandoned striker with Silva just abaft him. Silva revelled in the axial role, hitting a deflected opener four account afterwards the restart and axis in a additional from abutting ambit on the hour.


Despite all his artistic accuracy and abstruse excellence, Silva is not a acclaimed goalscorer but he enjoyed the role on Saturday.


"I accept consistently said that I like to play in the average and every time I play there I try to account goals, as if I play wide, but I adopt a axial position," the 28-year-old told City-limits TV. "I was able to account two goals and advice the team, so I am actual happy. Afterwards strikers we accept to plan a bit added on aggravating to account goals and we did that, as in the endure game, so hopefully we can abide like this."


Let's see if they bead some credibility and we can put some burden on them.David Silva, Manchester City-limits midfielder on Chelsea.


City trailed Chelsea by eight credibility at the end of endure ages but aggressiveness is now aerial afterwards a run of eight alternating wins in all competitions. Attention now accouterment to Chelsea's acknowledgment as they play their bold in duke at Stoke on Monday.


"We apperceive that we accept to play a lot of amateur during this ages and that we accept to try to get as abounding credibility as we can," Silva said. "So let's see if they bead some credibility and we can put some burden on them."


City bedeviled ascendancy throughout adjoin Palace but the visitors, for whom Yannick Bolasie impressed, had some acceptable opportunities in a aimless aboriginal half, conspicuously if Fraizer Campbell put a close-range aerial bang wide. City-limits bootless to get any of their eight first-half attempts on ambition - with Pablo Zabaleta traveling the abutting - but stepped up a accessory afterwards the break.

来自 “ ITPUB博客 ” ,链接:http://blog.itpub.net/30024016/viewspace-1437189/,如需转载,请注明出处,否则将追究法律责任。

转载于:http://blog.itpub.net/30024016/viewspace-1437189/

### 含义 `dp[i][j - coins[i - 1]]` 通常出现在动态规划(Dynamic Programming)的场景中,`dp` 一般是一个二维数组,用于存储子问题的解。`i` 通常表示考虑前 `i` 个物品(这里是硬币),`j` 表示当前要凑出的金额。`coins[i - 1]` 表示第 `i` 个硬币的面额。所以 `dp[i][j - coins[i - 1]]` 表示使用前 `i` 个硬币,凑出金额 `j - coins[i - 1]` 时的最优解(如最少硬币数量、最大价值等)。 ### 用途 这个代码片段的主要用途是在动态规划的状态转移过程中。在处理硬币找零或者背包问题时,当考虑是否使用第 `i` 个硬币来凑出金额 `j` 时,会用到这个值。通过比较使用或不使用第 `i` 个硬币的情况,来更新 `dp[i][j]` 的值,从而得到最优解。例如在硬币找零问题中,若要凑出金额 `j`,可以选择不使用第 `i` 个硬币,此时解为 `dp[i - 1][j]`;也可以选择使用第 `i` 个硬币,此时解为 `dp[i][j - coins[i - 1]] + 1`(加 1 表示使用了一个第 `i` 个硬币),然后取两者中的最优值更新 `dp[i][j]`。 ### 应用场景 - **硬币找零问题**:给定不同面额的硬币 `coins` 和一个总金额 `amount`,计算凑成总金额所需的最少硬币个数。在这种场景下,`dp[i][j]` 表示使用前 `i` 个硬币凑出金额 `j` 所需的最少硬币数,`dp[i][j - coins[i - 1]]` 用于状态转移,帮助更新 `dp[i][j]` 的值。 - **0 - 1 背包问题**:有 `n` 个物品,每个物品有对应的重量 `weight` 和价值 `value`,以及一个容量为 `capacity` 的背包,求在不超过背包容量的情况下能获得的最大价值。这里 `dp[i][j]` 表示前 `i` 个物品放入容量为 `j` 的背包中能获得的最大价值,`dp[i][j - weight[i - 1]]` 类似 `dp[i][j - coins[i - 1]]`,用于状态转移更新 `dp[i][j]`。 ### 具体实现示例(硬币找零问题) ```java public class CoinChange { public int coinChange(int[] coins, int amount) { int n = coins.length; int[][] dp = new int[n + 1][amount + 1]; // 初始化第一行,表示不使用任何硬币,除了金额为 0 时所需硬币数为 0,其他都为无穷大 for (int j = 1; j <= amount; j++) { dp[0][j] = Integer.MAX_VALUE - 1; } for (int i = 1; i <= n; i++) { for (int j = 1; j <= amount; j++) { if (j < coins[i - 1]) { // 当前金额小于第 i 个硬币面额,不能使用该硬币 dp[i][j] = dp[i - 1][j]; } else { // 可以选择使用或不使用第 i 个硬币 dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - coins[i - 1]] + 1); } } } return dp[n][amount] == Integer.MAX_VALUE - 1 ? -1 : dp[n][amount]; } } ```
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