Subsequence
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 18795 Accepted: 8043
Description
A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.
Input
The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The input will finish with the end of file.
Output
For each the case the program has to print the result on separate line of the output file.if no answer, print 0.
Sample Input
2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5
Sample Output
2
3
解题心得：
- 题意就是给你ｎ个数，要你选择连续的一段区间，其总和不小于ｓ，问你最短的区间长度是多少。
- 尺取法的模板题，尺取法很有意思，也被称之为尺取虫，这个可以说是很形象了，可以看成是一个一定长度的虫在线性表中爬动，复杂度就是O(N)。关于尺取有很多种写法，不同的写法复杂度不同，这里讲三种写法
- 第一种就是二分加尺取法，首先二分出一个长度，然后用尺取法去数组中检验这个二分出来的总和，复杂度为O(NlogN)。
- 第二种感觉和尺取关系不大，用的是二分，先求出前缀和，以每一个前缀和为起点，当前前缀和+ｓ为终点，因为前缀和是一个递增的数列，所以就可以二分查找终点，然后得到长度。复杂度为O(NlogN)。
- 第三种就是尺取，只不过这个尺取虫的长度是可以变换的，我们找一个最小的长度就行了，从起点开始每次保证尺取范围的总和不小于ｓ，同时向后面移动。复杂度为O(N).
第一种尺取代码
#include <algorithm>
#include <iostream>
#include <stdio.h>
using namespace std;
const int maxn = 1e5+10;
int num[maxn],n,s;
void init() {
scanf("%d%d",&n,&s);
for(int i=0;i<n;i++)
scanf("%d",&num[i]);
}
bool checke(int len) {
int sum = 0;
for(int i=0;i<len;i++)
sum += num[i];
if(sum >= s)
return true;
int l = 0,r = len-1;
while(r < n) {//
sum -= num[l];
l++,r++;
sum += num[r];
if(sum >= s)
return true;
}
return false;
}
int binary_search() {//二分尺取虫的长度
int l = 1 ,r = n;
while(r - l > 1) {
int mid = (l + r) >> 1;
if(checke(mid))
r = mid;
else
l = mid;
}
return r;
}
bool check_ans() {
int sum = 0;
for(int i=0;i<n;i++)
sum += num[i];
if(sum < s)
return true;
return false;
}
int main() {
int t;
scanf("%d",&t);
while(t--) {
init();
if(check_ans()) {//如果总和都小于ｓ直接输出０
printf("0\n");
continue;
}
int ans = binary_search();
printf("%d\n", ans);
}
return 0;
}第二种写法代码：
#include <algorithm>
#include <iostream>
#include <stdio.h>
using namespace std;
const int maxn = 1e5+10;
int sum[maxn],n,s;
void init() {
scanf("%d%d",&n,&s);
for(int i=1;i<=n;i++) {
int temp;
scanf("%d",&temp);
sum[i] = sum[i-1]+temp;
}
}
int solve() {
int ans = 0x7f7f7f7f;
for(int i=0;sum[i]+s<=sum[n];i++) {
int last = sum[i] + s;
int pos = lower_bound(sum+i,sum+n+1,last) - sum;//就是一个二分查找
int len = pos - i;
ans = min(ans,len);
}
return ans;
}
int main() {
int t;
scanf("%d",&t);
while(t--) {
init();
int ans = solve();
if(sum[n] < s)
ans = 0;
printf("%d\n",ans);
}
return 0;
}第三种写法代码：
#include <algorithm>
#include <iostream>
#include <stdio.h>
using namespace std;
const int maxn = 1e5+10;
int n,s,num[maxn];
int main() {
int t;
scanf("%d",&t);
while(t--) {
int sum,l,r,ans;
sum = l = r = 0;
ans = 0x7f7f7f7f;
scanf("%d%d", &n, &s);
for (int i = 0; i < n; i++)
scanf("%d", &num[i]);
while (1) {
while (sum < s && r < n) {
sum += num[r++];
}
if (sum < s)
break;
ans = min(ans, r - l);
sum -= num[l++];
}
if (ans > n)
ans = 0;
printf("%d\n",ans);
}
return 0;
}