LA 2678 Subsequence

题意：给出n个整数，给定整数S,求长度最短的连续序列，使它们的和大于或等于S

第一种，是求前缀和，然后用lower_bound来找

第二种，尺取法，

 1 #include<iostream>  

 2 #include<cstdio>  

 3 #include<cstring> 

 4 #include <cmath> 

 5 #include<stack>

 6 #include<vector>

 7 #include<map> 

 8 #include<set>

 9 #include<queue> 

10 #include<algorithm>  

11 using namespace std;

12 

13 typedef long long LL;

14 const int INF = (1<<30)-1;

15 const int mod=1000000007;

16 const int maxn=100005;

17 

18 int n,S;

19 int a[maxn],sum[maxn];

20 

21 int main(){

22     while(scanf("%d %d",&n,&S) != EOF){

23         memset(a,0,sizeof(a));

24         memset(sum,0,sizeof(sum));

25         for(int i = 0;i < n;i++) scanf("%d",&a[i]);

26         sum[0] = a[0];

27         for(int i = 1;i < n;i++) sum[i] = sum[i-1] + a[i];

28         

29     //    for(int i =0;i <= n;i++)

30     //    printf("sum[%d] = %d\n",i,sum[i]);

31         if(sum[n-1] < S){

32             printf("0\n");

33         }

34         else{

35             int res = INF;

36         for(int s = 0;sum[s] + S <= sum[n-1];s++){

37             int t = lower_bound(sum,sum + n,sum[s] + S) - sum;

38             res = min(res,t-s);

39         //    printf("res = %d\n",res);

40         }

41         printf("%d\n",res);

42         }        

43     }

44     return 0;

45 }

View Code

 

 1 #include<iostream>  

 2 #include<cstdio>  

 3 #include<cstring> 

 4 #include <cmath> 

 5 #include<stack>

 6 #include<vector>

 7 #include<map> 

 8 #include<set>

 9 #include<queue> 

10 #include<algorithm>  

11 using namespace std;

12 

13 typedef long long LL;

14 const int INF = (1<<30)-1;

15 const int mod=1000000007;

16 const int maxn=100005;

17 

18 int a[maxn];

19 int n,S;

20 

21 void solve(){

22     int t = 0,s = 0;

23     int res = INF;

24     int sum = 0;

25     for(;;){

26         while(t < n && sum < S){

27             sum += a[t++];

28         }

29         if(sum < S) break;

30         res = min(res,t - s);

31         sum -= a[s++];

32     }

33     if(res > n) res = 0;

34     printf("%d\n",res);

35 }

36 

37 int main(){

38     while(scanf("%d %d",&n,&S) != EOF){

39         memset(a,0,sizeof(a));

40         for(int i = 0;i < n;i++) scanf("%d",&a[i]);

41         solve();

42     }

43     return 0;

44 }

View Code