题解：用二分查询一下每次满足长度的下一个加上它的长度。

#include 
    
     using namespace std;typedef long long ll;ll a[500005];ll b[500005] = {0};int main(){    ll n,i,j,k,p=1,l,r;    char c;    memset(a,0,sizeof(a));    while((c=getchar()) != '\n')    {        if(c == ' '){a[p]++;p++;}        else a[p]++;    }    a[p]++;    for(i = 1; i <= p; i++)b[i] = b[i - 1] + a[i];    ll sum,ans;    scanf("%lld%lld",&l,&r);    for(k=l + 1;k<=r + 1;k++)    {        ans = 0;        int qwe = 0;        for(i=1;i<=p;)        {            if(i == 1)            {                ans += a[1] - 1;                qwe++;                i = upper_bound(b + 1, b + p + 1, k + b[i - 1]) - (b);                ans += a[i] - 1;                qwe++;            }            else            {                i = upper_bound(b + 1, b + p + 1, k + b[i - 1]) - (b);                ans += a[i] - 1;                qwe++;            }        }        printf("%lld\n",ans + qwe - 1);    }    return 0;}