URAL 1995 Illegal spices

构造。

前$n-k$个都是$1$，最后$k$个进行构造，首先选择填与上一个数字一样，如果不可行，那么这一格的值$+1$。

#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define ms(x,y) memset(x,y,sizeof(x))
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])
#define inone(x) scanf("%d",&x)
#define intwo(x,y) scanf("%d%d",&x,&y)
#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p)
#define lson x<<1,l,mid
#define rson x<<1|1,mid+1,r
#define mp(i,j) make_pair(i,j)
#define ft first
#define sd second
typedef long long LL;
typedef pair<int, int> pii;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
const int M = 1e4 + 1;
const double eps = 1e-10;
int T, n ,m, k, p;

int c[100010];
int a[100010];
long long s;

void update(int x,int val)
{
    while(x<=100000)
    {
        c[x]=c[x]+val;
        x=x+low(x);
    }
}

int sum(int x)
{
    int res=0;
    while(x>0)
    {
        res=res+c[x];
        x=x-low(x);
    }
    return res;
}

int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        scanf("%d",&p); s=0;

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

        int now=2,x=0,cnt=n-k;

        for(int i=n-k+1;i<=n;i++)
        {
            a[i] = now;
            if(cnt*100>=p*(i-1))
            {
                x++;
            }

            else
            {
                now++;
                cnt+=x;
                a[i]=now;
                x=1;
            }
        }

        for(int i=1;i<=n;i++) s=s+(long long) a[i];

        printf("%lld
",s);
        for(int i=1;i<=n;i++)
        {
            printf("%d",a[i]);
            if(i<n) printf(" ");
            else printf("
");
        }
    }
    return 0;
}