原题链接:poj3061 Subsequence
题意:给定长度为n的数列a[0]....a[n-1]以及S,求总和不小于S的连续子序列的长度的最小值,
若不存在,输出0
//尺取加二分 复杂度O(nlog n)
#include <cstdio>
const int MAX_N = 100010;
int N, S;
int a;
int sum[MAX_N];
bool judge(int x){
for(int i = x;i <= N;i ++){
if(sum[i] - sum[i - x] >= S)
return true;
}
return false;
}
void solve(){
if(sum[N] < S){
printf("0\n");
return;
}
int l = 0, r = N;
while(l <= r){ //二分查找
int mid = (l + r) / 2;
if(judge(mid))
r = mid - 1;
else
l = mid + 1;
}
printf("%d\n", l);
}
int main(){
int T;
scanf("%d", &T);
while(T --){
scanf("%d%d", &N, &S);
for(int i = 0;i < N;i ++){
scanf("%d", &a);
sum[i + 1] = sum[i] + a;
}
solve();
}
return 0;
}
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAX_N = 100010;
int N, S;
int sum[MAX_N];
void solve(){
if(sum[N] < S){
printf("0\n");
return;
}
int res = N;
for(int s = 0;sum[s] + S <= sum[N];s ++){ //调用STL库函数中的二分查找
int t = lower_bound(sum + s, sum + N, sum[s] + S) - sum;
res = min(res, t - s);
}
printf("%d\n", res);
}
int main(){
int T, a;
scanf("%d", &T);
while(T --){
scanf("%d%d", &N, &S);
for(int i = 0;i < N;i ++){
scanf("%d", &a);
sum[i + 1] = sum[i] + a;
}
solve();
}
return 0;
}
//尺取法:反复推进区间的开头和末尾,来求取满足条件的最小区间的方法
//复杂度O(n)
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAX_N = 100010;
int N, S;
int a[MAX_N];
void solve(){
int res = N + 1;
int s = 0, t = 0, sum = 0;
while(true){
while(t < N && sum < S)
sum += a[t ++];
if(sum < S) break;
res = min(res, t - s);
sum -= a[s ++];
}
if(res > N) res = 0;
printf("%d\n", res);
}
int main(){
int T;
scanf("%d", &T);
while(T --){
scanf("%d%d", &N, &S);
for(int i = 0;i < N;i ++)
scanf("%d", &a[i]);
solve();
}
return 0;
}