标签：tmp har fine while div unsigned long double style
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
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\n%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\n%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\n%d", tmp); return 0;} } } } puts("NO"); return 0; } /* 2 2 1 1 1 2 1 1 */
cf550C. Divisibility by Eight(结论)
标签：tmp har fine while div unsigned long double style
原文地址：https://www.cnblogs.com/zwfymqz/p/9571183.html
