牛客练习赛13
题解：出现了最多次数的只可能是4或者7
1 #pragma warning(disable:4996) 2 #include<queue> 3 #include<cstdio> 4 #include<bitset> 5 #include<cstring> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 10 int cnt1, cnt2, len; 11 12 int main() 13 { 14 char s[100]; 15 scanf("%s", s); 16 len = strlen(s); 17 cnt1 = cnt2 = 0; 18 for (int i = 0; i < len; i++) { 19 if (s[i] == '4') cnt1++; 20 if (s[i] == '7') cnt2++; 21 } 22 if (cnt1 == 0 && cnt2 == 0) printf("-1\n"); 23 else if (cnt1 >= cnt2) printf("4\n"); 24 else printf("7\n"); 25 }
题解：先构造出在1e10范围内的幸运数，然后分段讨论。
1 #pragma warning(disable:4996) 2 #include<queue> 3 #include<cstdio> 4 #include<bitset> 5 #include<cstring> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 typedef long long ll; 10 11 const int maxn = 2000; 12 13 ll cnt = 0; 14 ll a[maxn]; 15 16 void DFS(ll x, int d) { 17 a[cnt++] = x; 18 if (d == 9) return; 19 DFS(x * 10 + 4, d + 1); 20 DFS(x * 10 + 7, d + 1); 21 } 22 23 void Inite() { 24 DFS(0, 0); 25 sort(a, a + cnt); 26 a[cnt++] = 4444444444; 27 } 28 29 int main() 30 { 31 Inite(); 32 ll l, r; 33 while (scanf("%lld%lld", &l, &r)!=EOF) { 34 ll ans = 0, pos = 0; 35 while (l <= r) { 36 while (a[pos] < l) pos++; 37 ans += (min(a[pos], r) - l + 1)*a[pos]; 38 l = a[pos] + 1; 39 } 40 printf("%lld\n", ans); 41 } 42 }
题解：注意到两种情况，447，4在偶数的位置上。477，4在奇数的位置上。剩下的直接改就行。
1 #pragma warning(disable:4996) 2 #include<queue> 3 #include<cstdio> 4 #include<bitset> 5 #include<cstring> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 10 const int maxn = 1e5 + 5; 11 12 int n, k; 13 char s[maxn]; 14 15 int main() 16 { 17 while (scanf("%d%d", &n, &k) != EOF) { 18 scanf("%s", s + 1); 19 int l = strlen(s+1); 20 for (int i = 1; i < l; i++) { 21 if (s[i] == '4' && s[i + 1] == '7') { 22 //if (i == 1) { s[i + 1] = '4'; k--; continue; } 23 if (i & 1) { 24 if (i + 1 != l && s[i + 2] == '7') { 25 if (k & 1) s[i + 1] = '4'; 26 break; 27 } 28 else s[i + 1] = '4'; 29 } 30 else { 31 if (i - 1 != 0 && s[i - 1] == '4') { 32 if ((k & 1)) s[i] = '7'; 33 break; 34 } 35 else s[i] = '7'; 36 } 37 k--; 38 } 39 } 40 printf("%s\n", s + 1); 41 } 42 return 0; 43 }
