E. Good Triples

合集 - codeforces(31)

1.B. Rudolf and 1212024-03-162.C. Rudolf and the Ugly String2024-03-163.A. Rudolf and the Ticket2024-03-164.D. Rudolf and the Ball Game2024-03-185.G. Rudolf and Subway2024-04-066.C. Choose the Different Ones!2024-02-097.B. Following the String2024-02-098.E. Klever Permutation2024-02-099.D. Find the Different Ones!2024-02-0910.B. YetnotherrokenKeoard2023-12-1011.C. Removal of Unattractive Pairs2023-12-1012.D. Jumping Through Segments2023-12-11

13.E. Good Triples2023-12-13

14.F. Shift and Reverse2023-12-1315.C. Nikita and LCM2024-05-2916.B. Prefiquence2024-05-0317.E. Vlad and a Pair of Numbers2024-04-0718.D. XOR Construction2024-03-2719.C. Non-coprime Split2024-01-2720.D. Divide and Equalize2024-06-0221.E. Block Sequence2024-06-0222.D. In Love2024-06-1123.C. Theofanis' Nightmare2024-06-2224.D. Yet Another Monster Fight2024-06-2225.C. Count Triangles2024-07-1126.E. Count Paths2024-07-1327.Collecting Numbers II2024-07-3128.C. Even Subarrays2024-08-0729.C. Perform Operations to Maximize Score2024-08-2030.F. Color Rows and Columns2024-08-1631.C. To Become Max2024-09-10

收起

首先假定已经找到abc符合题目条件。则我们假定a1,a2,a3...;b1,b2,b3...;c1,c2,c3...为abc各个位置上的数字，那么

   a1  a2  a3

   b1  b2  b3

+ c1  c2  c3

----------------

   x1   x2   x3

又由digsum等式可知a1+b1+c1+...=x1+x2+x3。

那么我们根据竖式不难发现在每一位进行加法时不能够产生进位，否则将不满足digsum等式。

即a3+b3+c3=x3;a2+b2+c2=x2;a1+b1+c1=x1。

所以我们只需要求出x每一位有几种拆法，然后再相乘。

拆分方法我们可以借鉴高中隔板法：有Xi个小球要求分成三部分，有几种分法。

主要代码：

#include<bits/stdc++.h>
using namespace std;
int main(){
    int a[15]={1,3,6,10,15,21,28,36,45,55};  //Xi个小球的分法 
    int t;
    cin>>t;
    while (t--){
        int n;
        cin>>n;
        if (n==0){   //0需要特判 
            printf("%d\n",1);
            continue;
        }
        long long int sum=1;  //longlongint避免溢出 
        while (n!=0){   //每一位方法数相乘 
            sum*=a[n%10];
            n/=10;
        }
        cout<<sum<<endl;
    }
    return 0;
}