B. Worms
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题目来源
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such thati-th pile containsai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers1 toa1, worms in second pile are labeled with numbersa1 + 1 toa1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103,a1 + a2 + ... + an ≤ 106), where ai is the number of worms in thei-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Print m lines to the standard output. Thei-th line should contain an integer, representing the number of the pile where the worm labeled with the numberqi is.
52 7 3 4 931 25 11
153
For the sample input:
- The worms with labels from [1, 2] are in the first pile.
- The worms with labels from [3, 9] are in the second pile.
- The worms with labels from [10, 12] are in the third pile.
- The worms with labels from [13, 16] are in the fourth pile.
- The worms with labels from [17, 25] are in the fifth pile.
题意:n个数,每个数表示区间的长度,比如2表示区间【1,2】为第一段, 7表示区间【3,9】为第二段,后面n-2个数以此类推。
m次访问,比如,1, 25, 11的意思是,问1是第几段的,25是第几段的,11是第几段的。
思路:用一个数组标记记录,比如[3,9]是第二段,让a[3]=2,a[4]=2,a[5]=2,...a[8]=,a[9]=2.这样给定一个数m,只要求出a[m]的值就知道它在哪一段了。
代码:
#include<cstdio>int s[1000010];using namespace std;int main(){ int n,t,a=1,k=1,m; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&t); for(int j=a;j<t+a;j++) s[j]=k; a+=t; k++; } scanf("%d",&m); for(int i=0;i<m;i++) { scanf("%d",&t); printf("%d\n",s[t]); } return 0;}
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