A. Tricky Sum Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/598/problem/A Description In this problem you are to calculate the sum of all integers from 1 to n, but you should take all powers of two with minus in the sum.…

Educational Codeforces Round 53 E. Segment Sum 题意: 问[L,R]区间内有多少个数满足:其由不超过k种数字构成. 思路: 数位DP裸题,也比较好想.由于没考虑到前导0,卡了很久.但最惨的是,由于每次求和的时候需要用到10的pos次幂,我是用提前算好的10的最高次幂,然后每次除以10往下传参.但我手贱取模了,导致每次除以10之后答案就不同余了,这个NC细节错误卡了我一小时才发现. 代码: #include<iostream> #include<…

F. The Sum of the k-th Powers 题目连接: http://www.codeforces.com/contest/622/problem/F Description There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees. Find the value of the sum modulo 109 + 7 (so you shoul…

A. Tennis Tournament 题目连接: http://www.codeforces.com/contest/628/problem/A Description A tennis tournament with n participants is running. The participants are playing by an olympic system, so the winners move on and the losers drop out. The tourname…

题意 给定序列$a_n$,每次将$[L,R]$区间内的数$a_i$替换为$d(a_i)$,或者询问区间和 这题和区间开方有相同的操作 对于$a_i \in (1,10^6)$,$10$次$d(a_i)$以内肯定可以最终化为$1$或者$2$,所以线段树记录区间最大值和区间和,$Max\le2$就返回,单点暴力更新,最后线性筛预处理出$d$ 时间复杂度$O(m\log n)$ 代码 #include <bits/stdc++.h> using namespace std; typedef long…

The Sum of the k-th Powers There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees. Find the value of the sum modulo 109 + 7 (so you should find the remainder after dividing the answer by the value 109 + 7).…

FallDream dalao找的插值练习题 题目大意:给定n,k,求Σi^k (i=1~n),对1e9+7取模.(n<=10^9,k<=10^6) 思路:令f(n)=Σi^k (i=1~n),则有f(n)-f(n-1)=n^k,说明f(n)的差分是n的k次多项式,则所求f(n)为n的k+1次多项式,利用拉格朗日插值公式,我们暴力计算n=0~k+1时的答案,代入公式,利用预处理的信息加速计算,总复杂度O(klogMOD). #include<cstdio> #define MOD…

B. New Skateboard 题目连接: http://www.codeforces.com/contest/628/problem/A Description Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask som…

传送门 题意 给出n个数,q个询问,每个询问有两个数p,k,询问p+k+a[p]操作几次后超过n 分析 分块处理,在k<sqrt(n)时,用dp,大于sqrt(n)用暴力 trick 代码 #include<cstdio> int n,a[100100],p,k,q,dp[100100][350]; int main() { scanf("%d",&n); for(int i=1;i<=n;++i) scanf("%d",a+i);…

You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist. Divisor of n is any such natural number, that n can be divided by it without remainder. Input The first line contains two integers n and k (1 ≤ n ≤ …