E. Final Countdown

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-062.C. Removal of Unattractive Pairs2023-12-063.D. Jumping Through Segments2023-12-084.E. Good Triples2023-12-085.F. Shift and Reverse2023-12-086.D. Yet Another Monster Fight2023-12-117.A. Constructive Problems2023-12-188.C. Game with Multiset2023-12-199.A. Rating Increase2023-12-1910.B. Swap and Delete2023-12-1911.A. Problemsolving Log2023-12-2012.B. Preparing for the Contest2023-12-2013.C. Quests2023-12-2014.D. Three Activities2023-12-2015.E2. Game with Marbles (Hard Version)2023-12-2016.cf刷题有感2023-12-2017.A. Anonymous Informant2023-12-2118.A. Forked!2023-12-2319.B. Make Almost Equal With Mod2023-12-2720.C. Heavy Intervals2023-12-2721.D. Split Plus K2023-12-2822.A. 20232023-12-3123.B. Two Divisors2023-12-3124.C. Training Before the Olympiad2023-12-3125.D. Mathematical Problem2024-01-0226.F. Greetings2024-01-0227.C. Partitioning the Array2024-01-1528.G. Bicycles2024-01-1629.E. Eat the Chip2024-01-1730.G. Lights2024-01-1731.D. Array Repetition2024-01-1732.D. Berserk Monsters2024-01-2233.E. Increasing Subsequences2024-01-2334.D. Very Different Array2024-01-2435.G. Mischievous Shooter2024-01-2536.B. Plus-Minus Split2024-01-2737.B. A Balanced Problemset2024-01-2838.C. Did We Get Everything Covered2024-01-2839.D. Find the Different Ones!2024-02-0840.C. Grouping Increases2024-02-1041.D. Good Trip2024-02-1242.C. Physical Education Lesson2024-02-12

43.E. Final Countdown2024-02-18

44.D. Divisible Pairs2024-02-1945.G. Vlad and Trouble at MIT2024-02-2046.A. Brick Wall2024-02-2847.B. Minimize Inversions2024-02-2848.C. XOR-distance2024-02-2849.A. Moving Chips2024-02-2850.B. Monsters Attack!2024-02-2851.C. Find B2024-02-2852.D. Slimes2024-02-2853.C. Turtle Fingers: Count the Values of k2024-02-2854.D. Turtle Tenacity: Continual Mods2024-02-2855.D. Vlad and Division2024-02-2856.C. LR-remainders2024-02-2857.D. Lonely Mountain Dungeons2024-03-12

收起

原题链接

题解

本题中，每一位数字的每一次变化都会对答案贡献1，所以对于第 $i$ 位数字而言，它的贡献为从最左边到现在的数，设为 $f[i]$

所以答案为 $\sum _{i=1}^{n}f[i]$，可以用高精度加法解决
然而这样一来时间复杂度就超了 $O(t·n^2)$

所以我们尝试从中找寻一些规律
我们发现，第 $i$ 的数字会对答案 $[1,n-i+1]$ 位上的数字贡献加 $s[i]$ ，
而我们统计每一位数字又是线性的， 即我们在统计完第 $i$ 位的数字时，还要再统计 $n-i$ 个数字
所以我们可以在 $O(t·n)$ 的时间内完成

code

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int a[400005]={0};
        int n;
        cin >> n;
        char s[n+5];
        scanf("%s",s+1);
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            cnt+=s[i]-'0';
            a[n-i+1]=cnt;
        }

        for(int i=1;i<=n;i++)
        {
            a[i+1]+=a[i]/10;
            a[i]%=10;
        }

        int start=n+1;
        if(a[start])
        {
            while(a[start])
            {
                a[start+1]=a[start]/10;
                a[start]%=10;
                start++;
            }
            start--;
        }
        else
        {
            while(!a[start])start--;
        }
        for(int i=start;i>=1;i--)cout<<a[i];
        puts("");
    }

    return 0;
}