LA 2678 Subsequence
2018-02-04 19:03
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题目:Subsequence
思路:使用前缀和优化。枚举终点 j,计算出起点 i。由于数组a递增,所以当 j1 > j2 时,一定有 i1 > i2 。
代码:
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <cstring>
#include <map>
#include <cmath>
using namespace std;
#define maxn 100000
#define ll long long
int n,s;
ll a[maxn+5];
int main() {
while(~scanf("%d%d",&n,&s)) {
for(int i=1; i<=n; i++) {
scanf("%lld",&a[i]);
a[i]+=a[i-1];
}
int i=1;
int ans=n+1;
for(int j=1; j<=n; j++) {
if(a[j]-a[i-1]<s) continue;
while(a[j]-a[i]>=s) {
i++;
}
ans=min(ans,j-i+1);
}
if(ans==n+1) printf("0\n");
else printf("%d\n",ans);
}
return 0;
}
思路:使用前缀和优化。枚举终点 j,计算出起点 i。由于数组a递增,所以当 j1 > j2 时,一定有 i1 > i2 。
代码:
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <cstring>
#include <map>
#include <cmath>
using namespace std;
#define maxn 100000
#define ll long long
int n,s;
ll a[maxn+5];
int main() {
while(~scanf("%d%d",&n,&s)) {
for(int i=1; i<=n; i++) {
scanf("%lld",&a[i]);
a[i]+=a[i-1];
}
int i=1;
int ans=n+1;
for(int j=1; j<=n; j++) {
if(a[j]-a[i-1]<s) continue;
while(a[j]-a[i]>=s) {
i++;
}
ans=min(ans,j-i+1);
}
if(ans==n+1) printf("0\n");
else printf("%d\n",ans);
}
return 0;
}
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