CF1157D N Problems During K Days

思路

简单的贪心
序列肯定是由几段公差为1的等差数列组成的，然后之间计算剩下的数字能有多大的和，不够的话整体上升1补上就好了

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#define int long long
using namespace std;
int n,k,last=0,a[101000];
int sum(int first,int times){
    return (first+times-1+first)*times/2;
}
signed main(){
    scanf("%d %d",&n,&k);
    for(int i=1;i<=k;i++){
        int q=last+1,t=k-i+1;
        int delta=max(n-sum(q,t),0LL);
        if(i!=1)
            q+=min(last-1,delta/t);
        else
            q+=delta/t;
        n-=q;
        last=q;
        a[i]=q;
    }
    if(n!=0)
        printf("NO
");
    else{
        printf("YES
");
        for(int i=1;i<=k;i++)
            printf("%d ",a[i]);
        printf("
");
    }
    return 0;
}