第二次集訓——2011年2月27日

      很僥倖，昨天遲迴去，隊長剛好沒在，躲過了一劫，但是要牢記，不能超假啊，後果很嚴重的。

      今天感到了集訓的自由性~~上午還是做昨天的題目，有點反感，就去做09年的校賽題目，發現一道很好的題目，楊輝三角！

      題目大意是給出在楊輝三角中的相鄰兩個數的位置，求出在這個圖形中的位置。

      題目數據是給出的數在LONG INT範圍內，可是具體行列到哪種程度不敢想象~~~

      當時做的人基本都是混過去的，就是說默認在INT64範圍內做，無法顧及到很多數據，所以題目的數據弱，竟然都過了~~~

      我想到了解方程的方法，根據兩個值列兩個方程，但是很顯然這是個有多個解的方程，解還得去驗證到底對不對，中間試解的過程比較慢，但由於題目數據的弱，行數不會超過50行，所以試根也很快了，但是遇到犀利一點的數據怎麼辦呢？比如99999999 1

      恰巧剛看過擴展的歐幾里得算GCD的方法，有了大概印象，可以求一個一元二次方程整數解，然後就可以把通解表示出來，這道題需要這個，這樣一次就可以得一個根，速度接近線性了！爭取在兩天內把它編寫出來！

 

 

      下午是個鬱悶的下午。題目都是很噁心的，描述一大堆，做做出了2題。

A題：好像是跟NOIP差不多的題目

算法：二分+貪心

         由於要使得最小的使用度儘量大，二分使用度；每樣物品必須要出現購買一樣，這樣就可以購買使用度符合範圍的，且MONEY儘量少的，就可以判斷當前解是否可行；

注意了，二分的端點很重要的，這裏左端點可以選擇0，右端點選擇出現過的最大使用度。呃~~~Wrong了，應該要加1，因爲左邊是可以取到的，右邊不可取到，這樣要考慮到最好的情況：最後的答案剛好是最大的使用度。

Description Recently your team noticed that the computer you use to practice for programming contests is not good enough anymore. Therefore, you decide to buy a new computer. To make the ideal computer for your needs, you decide to buy separate components and assemble the computer yourself. You need to buy exactly one of each type of component. The problem is which components to buy. As you all know, the quality of a computer is equal to the quality of its weakest component. Therefore, you want to maximize the quality of the component with the lowest quality, while not exceeding your budget. Input On the first line one positive number: the number of testcases, at most 100. After that per testcase: • One line with two integers: 1 ≤ n ≤ 1 000, the number of available components and 1 ≤ b ≤ 1 000 000 000, your budget. • n lines in the following format: “type name price quality”, where type is a string with the type of the component, name is a string with the unique name of the com- ponent, price is an integer (0 ≤ price ≤ 1 000 000) which represents the price of the component and quality is an integer (0 ≤ quality ≤ 1 000 000 000) which represents the quality of the component (higher is better). The strings contain only letters, digits and underscores and have a maximal length of 20 characters. It will always possible to construct a computer with your budget. Output Per testcase: • One line with one integer: the maximal possible quality. Sample Input 118 800processor 3500_MHz 66 5processor 4200_MHz 103 7processor 5000_MHz 156 9processor 6000_MHz 219 12memory 1_GB 35 3memory 2_GB 88 6memory 4_GB 170 12mainbord all_onboard 52 10harddisk 250_GB 54 10harddisk 500_FB 99 12casing midi 36 10monitor 17_inch 157 5monitor 19_inch 175 7monitor 20_inch 210 9monitor 22_inch 293 12mouse cordless_optical 18 12mouse microsoft 30 9keyboard office 4 10 Sample Output 9

 

代碼：

#include <stdio.h> #include <string.h> const int maxn = 1010; int price[maxn], quality[maxn]; char name[maxn][22], kind[22], type[22]; int head[maxn], next[maxn], tot; int check() { int i, j; for (i = 1; i <= tot; i++) { bool flag = true; if (strlen(name[i]) != strlen(kind)) flag = false; else for (j = 0; j < strlen(kind); j++) if (name[i][j] != kind[j]) { flag = false; break; } if (flag) return i; } tot++; memcpy(name[tot],kind,sizeof(kind)); return tot; } int main() { int cases; scanf("%d", &cases); for (;cases>0; cases--) { int n, money, i, j; scanf("%d %d/n", &n, &money); tot = 0;//from 1.. memset(head,-1,sizeof(head)); int left = 0, right = 0; for (i = 0; i < n; i++) { scanf("%s %s %d %d/n", kind, type, &price[i], &quality[i]); if (quality[i] > right) right = quality[i]; int now = check(); next[i] = head[now]; head[now] = i; } //for (i = 1; i <= tot; i++) printf("%s/n", name[i]); //printf("%d/n",tot); right++;//notice that right can't be reached so right should be bigger than the maximal quality while (left+1 < right) { int mid = (left+right)/2, nowcost = 0; //printf("mid%d/n", mid); for (i = 1; i <= tot; i++) { int mincost = 1 << 30; for (j = head[i]; j != -1; j = next[j]) if (quality[j] >= mid && price[j] < mincost) mincost = price[j]; nowcost += mincost; if (mincost >= 1 << 30) break; } if (nowcost <= money) left = mid; else right = mid; } printf("%d/n", left); } return 0; }