無代碼
題目描述
Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken off the chain one by one. Once a diamond is off the chain, it cannot be taken back. For example, if we have a chain of 8 diamonds with values M$3, 2, 1, 5, 4, 6, 8, 7, and we must pay M$15. We may have 3 options:
- Cut the chain between 4 and 6, and take off the diamonds from the position 1 to 5 (with values 3+2+1+5+4=15).
- Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).
- Cut before 8, and take off the diamonds from the position 7 to 8 (with values 8+7=15).
Now given the chain of diamond values and the amount that a customer has to pay, you are supposed to list all the paying options for the customer.
If it is impossible to pay the exact amount, you must suggest solutions with minimum lost.
題意
給出一個數字序列與一個數S，在數字序列中求出所有和值爲S的連續子序列（區間下標左端點小的先輸出，左端點相同時右端點小的先輸出）。若沒有這樣的序列，求出和值恰好大於S的子序列（即在所有和值大於S的子序列中和值最接近S)。假設序列下標從1開始。
輸入格式
Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤105), the total number of diamonds on the chain, and M (≤108), the amount that the customer has to pay. Then the next line contains N positive numbers D1⋯DN (Di≤103 for all i=1,⋯,N) which are the values of the diamonds. All the numbers in a line are separated by a space.
輸出格式
For each test case, print i-j in a line for each pair of i ≤ j such that Di + … + Dj = M. Note that if there are more than one solution, all the solutions must be printed in increasing order of i.
If there is no solution, output i-j for pairs of i ≤ j such that Di + … + Dj >M with (Di + … + Dj −M) minimized. Again all the solutions must be printed in increasing order of i.
It is guaranteed that the total value of diamonds is sufficient to pay the given amount.
輸入樣例1
16 15
3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13
輸出樣例1
1-5
4-6
7-8
11-11
輸入樣例2
5 13
2 4 5 7 9
輸入樣例2
2-4
4-5
題目信息
作者：CHEN, Yue
單位：浙江大學
代碼長度限制：16 KB
時間限制：300 ms
內存限制：64 MB
題目類型：二分-尋找有序數列第一個滿足某條件的元素的位置
分析
設 Sum[i] 表示 a[1] 倒 a[i] 的和，也就是 Sum[i] = a[1] + a[2] + a[3] + …… + a[i]。
因爲 a 序列都是正值，所以 Sum 是遞增的，這樣如果我們要判斷 第 c 個數到第 d 個數的和，只要算 Sum[d] - Sum[c]。
那麼這道題就轉換成了判斷 Sum[i] - Sum[j] 的值是否爲目標值或者接近目標值
我們可以從0開始：Sum[j] - Sum[0]，並在每次遍歷中判斷出 Sum[j] - Sum[i] 距離目標值的遠近：Sum[j] - Sum[i] - 目標值，每次取最小值
int target = 要支付的價格，也就是目標值
int min = Integer.MAX_VALUE; // 從a[i]加到a[j]的值 與 目標值的差
for(int i =0; i < Sum.length; i++){
尋找Sum數列中，第一個滿足 Sum[j]-Sum[i]>=target 的 j
與此同時，要記錄下這個j所算得的 Sum[j]-Sum[i] 距離 target 的距離，如果比當前的 min 小，就替換掉它
}
最後輸出所有的 Sum[j]-Sum[i]-target==min 的i和j