D. Single-use Stones
Examples 1: input 10 5 0 0 1 0 2 0 0 1 0 output 3 Examples 2: input 10 3 1 1 1 1 2 1 1 1 1 output 3
Title:
w represents the width of the river, l represents the farthest distance the frog can jump, and the w-1 element in the second line represents that there are a[i] stones at the place i from the river bank, a If a stone is stepped on twice, ask how many frogs can jump across the river at most
analyze:
Because the number of frogs that can be skipped at most is determined by the minimum number of footholds, his problem is actually in [i, i+l], i∈[l,w-1], what is the minimum number of stones
Personal code:
#define debug
#include<stdio.h>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<functional>
#include<iomanip>
#include<map>
#include<set>
#define pb push_back
#define dbg(x) cout<<#x<<" = "<<(x)<<endl;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>PLL;
typedef pair<int,ll>Pil;
const ll INF = 0x3f3f3f3f;
const ll inf=0x7fffffff;
const double eps=1e-8;
const int maxn =1e6;
const int N = 510;
const ll mod=1e9+7;
const ll MOD=1e9;
//------
//define
ll sum[maxn];
ll a[maxn];
int w,l;
//solve
void solve() {
while(cin>>w>>l){
ll ans=inf;
for(int i=1;i<w;i++){
cin>>a[i];
sum[i]=sum[i-1]+a[i];
}
for(int i=l;i<=w-1;i++){
ans=min(ans,sum[i]-sum[i-l]);
}
cout<<ans<<endl;
memset(sum,0,sizeof(sum));
}
}
//main
int main() {
ios_base::sync_with_stdio(false);
#ifdef debug
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
cin.tie(0);
cout.tie(0);
solve();
/*
#ifdef debug
fclose(stdin);
fclose(stdout);
system("out.txt");
#endif
*/
return 0;
}