• cf550C. Divisibility by Eight(结论)


    题意

    给出长度为$n$的字符串,判断是否能删除一些数后被$8$整除

    Sol

    神仙题啊Orz

    结论:

    若数字的后三位能被$8$整除,则该数字能被$8$整除

    证明

    设$x = 10000 * a_i + 1000 * a_{i - 1} + dots$

    发现大于$3$的位都会分解出$8$这个因数

    然后大力枚举三个位置即可

    /*
    */
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<map>
    #include<vector>
    #include<set>
    #include<queue>
    #include<cmath>
    //#include<ext/pb_ds/assoc_container.hpp>
    //#include<ext/pb_ds/hash_policy.hpp>
    #define Pair pair<int, int>
    #define MP(x, y) make_pair(x, y)
    #define fi first
    #define se second
    #define int long long 
    #define LL long long 
    #define ull unsigned long long 
    #define rg register 
    #define pt(x) printf("%d ", x);
    //#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
    //char buf[(1 << 22)], *p1 = buf, *p2 = buf;
    //char obuf[1<<24], *O = obuf;
    //void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
    //#define OS  *O++ = ' ';
    using namespace std;
    //using namespace __gnu_pbds;
    const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
    const double eps = 1e-9;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N;
    int a[MAXN];
    char s[MAXN];
    main() {
        scanf("%s", s + 1); 
        N = strlen(s + 1);
        for(int i = 1; i <= N; i++) a[i] = s[i] - '0';
        for(int i = 1; i <= N; i++) 
            if(a[i] % 8 == 0) {printf("YES
    %d", a[i]); return 0;}
        for(int i = 1; i <= N; i++) {
            for(int j = i + 1; j <= N; j++) {
                int tmp = a[i] * 10 + a[j];
                if(tmp % 8 == 0) {printf("YES
    %d", tmp); return 0;}
            }
        }
        for(int i = 1; i <= N; i++) {
            for(int j = i + 1; j <= N; j++) {
                for(int k = j + 1; k <= N; k++) {
                    int tmp = a[i] * 100 + a[j] * 10 + a[k];
                    if(tmp % 8 == 0) {printf("YES
    %d", tmp); return 0;}
                }
            }
        }
        puts("NO");
        return 0;
    }
    /*
    2 2 1
    1 1
    2 1 1
    */
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/9571183.html
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