C. Torn Lucky Ticket

合集 - 模拟(54)

1.G. Vlad and Trouble at MIT2024-02-202.P2899 [USACO08JAN] Cell Phone Network G2024-02-203.Watering an Array2024-02-234.P2055 [ZJOI2009] 假期的宿舍2024-03-085.T432579 大西洋船王2024-03-086.T432580 星云 hard ver.2024-03-087.P7072 [CSP-J2020] 直播获奖2024-03-098.E - Insert or Erase2024-03-109.D - String Bags2024-03-1010.P2671 [NOIP2015 普及组] 求和2024-03-1011.P1884 [USACO12FEB] Overplanting S2024-03-1012.F. Microcycle2024-03-1213.E. Increasing Subsequences__22024-03-1314.G. Path Prefixes2024-03-1415.F. Rudolf and Imbalance2024-03-1416.P1355 神秘大三角2024-03-2017.[ARC172A] Chocolate2024-03-2518.F. 0, 1, 2, Tree!2024-03-2919.E. Nearly Shortest Repeating Substring2024-03-2920.P1354 房间最短路问题2024-03-2921.P3258 [JLOI2014] 松鼠的新家2024-04-0222.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-0523.P1337 [JSOI2004] 平衡点 / 吊打XXX2024-04-0724.P3745 [六省联考 2017] 期末考试2024-04-0825.D. Inaccurate Subsequence Search2024-04-0926.E. Long Inversions2024-04-0927.C. Deep Down Below2024-04-0928.C. Rings2024-04-0929.F. Unfair Game2024-04-1030.C. Inversion Graph2024-04-1031.D. Cross Coloring2024-04-1232.D2. Set To Max (Hard Version)2024-04-1233.C. Array Game2024-04-1234.F. Alex's whims2024-04-1235.Queue Sort2024-04-1236.P10252 线性变换2024-04-1637.P10288 [GESP样题 八级] 区间2024-04-1638.P2161 [SHOI2009] 会场预约2024-04-1739.L2-001 紧急救援2024-04-1840.L2-002 链表去重2024-04-1841.L2-013 红色警报2024-04-1842.L2-014 列车调度2024-04-1943.L2-022 重排链表2024-04-1944.L2-023 图着色问题2024-04-1945.L2-039 清点代码库2024-04-1946.L2-034 口罩发放2024-04-1947.L3-037 夺宝大赛2024-04-2348.仓库规划2024-04-2349.因子化简2024-04-23

50.C. Torn Lucky Ticket2024-04-27

51.俄罗斯方块2024-05-1452.F. Cutting Game2024-05-2153.D. Prefix Permutation Sums2024-06-1754.P3304 [SDOI2013] 直径2024-07-15

收起

原题链接

题解

1.题目对 $i,j$ 没有限制，也就是说，i可以大于j，可以小于j，也可以等于j
2.找普适规律。不管谁和谁成功配对，要么两个数长度相等；要么前面的数长度大于后面的数，于是前面的数分后半部分出来；要么后面的数大于前面的数，于是后面的数分前半部分出来；

code

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

ll ten[7];
ll a[200005];
int main()
{
    ten[0]=1;
    for(ll i=1;i<=6;i++) ten[i]=ten[i-1]*10;

    ll n;
    cin>>n;
    ll ans=0;
    for(ll i=1;i<=n;i++) cin>>a[i];
    sort(a+1,a+1+n);


    map<ll,ll> dp[7];
    for(ll i=1;i<=n;i++)
    {
        ll x=a[i];
        ll sum=0,len=0,tem=x;
        while(tem)
        {
            sum+=tem%10;
            len++;
            tem/=10;
        }

        ans+=dp[len][sum]*2;//左右呼唤

        ll pre=0;
        for(ll j=len;j>=len/2+2;j--)//后面的数分前半部分出来
        {
            pre=pre+x/ten[j-1]%10;//是数字之和不是数字截取
            ans+=dp[(j-1)-(len-(j-1))][sum-pre-pre];
        }

        pre=0;
        for(ll j=1;j<=len/2-(1-len%2);j++)//前面的数分后半部分出来
        {
            pre=pre+x/ten[j-1]%10;
            ans+=dp[len-j-j][sum-pre-pre];
        }

        dp[len][sum]++;
    }


    cout<<ans+n;
    return 0;
}