A：签到题
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<int,int> pii; const int N = 105; const int M = 5e4+5; const LL Mod = 199999; #define rg register #define pi acos(-1) #define INF 1e9 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < ‘0‘ || c > ‘9‘){if(c == ‘-‘) f = -1;c = getchar();} while(c >= ‘0‘ && c <= ‘9‘){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } void print(int x){ if(x < 0){x = -x;putchar(‘-‘);} if(x > 9) print(x/10); putchar(x%10+‘0‘); } } using namespace FASTIO; void FRE(){ /*freopen("data1.in","r",stdin); freopen("data1.out","w",stdout);*/} string s[1005]; int cnt[30]; int main() { int ca;ca = read(); while(ca--) { memset(cnt,0,sizeof(cnt)); int n;n = read(); for(rg int i = 1;i <= n;++i) { cin >> s[i]; for(rg int j = 0;j < s[i].size();++j) cnt[s[i][j]-‘a‘]++; } int f = 0; for(rg int j = 0;j < 26;++j) if(cnt[j]%n != 0) f = 1; printf("%s\n",f ? "NO" : "YES"); } // system("pause"); }
B：考虑枚举底数c。
那么这样复杂度是cn级别，显然不对。
但是仔细看可以发现，但c 稍微大一点，n稍微大一点，c^i就会超过1e18了。
然后对于选底数为1.最大花费也就1e5*1e9 = 1e14左右。
那么我们可以去枚举C，当某个点的单个花费都已经超过1e14了，那么肯定没有1的时候优，那么就退出。
可以发现，这样在c*n保持很小时就会结束了。
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<int,int> pii; const int N = 1e5+5; const int M = 5e4+5; const LL Mod = 199999; #define rg register #define pi acos(-1) #define INF 1e18 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < ‘0‘ || c > ‘9‘){if(c == ‘-‘) f = -1;c = getchar();} while(c >= ‘0‘ && c <= ‘9‘){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } void print(int x){ if(x < 0){x = -x;putchar(‘-‘);} if(x > 9) print(x/10); putchar(x%10+‘0‘); } } using namespace FASTIO; void FRE(){ /*freopen("data1.in","r",stdin); freopen("data1.out","w",stdout);*/} LL a[N],b[N],ans = INF; int main() { int n;n = read(); for(rg int i = 1;i <= n;++i) a[i] = read(); sort(a+1,a+n+1); int f = 0; for(rg int low = 1;;low++) { LL ma = 1; b[1] = 1; for(rg int i = 2;i <= n;++i) { ma *= low; b[i] = ma; if(ma > 1e15){f = 1;break;} } if(f) break; LL tmp = 0; for(rg int i = 1;i <= n;++i) tmp += abs(a[i]-b[i]); ans = min(ans,tmp); } printf("%lld\n",ans); //system("pause"); }
C：考虑一种思路。
对于前n-1个数，加上(n-1)*a[i]。
那么次数第i个位置的数肯定是n*a[i]。
那么满足时n的倍数，可以直接-n*a[i]，然后我们再让a[n]加上n*a[n]-a[n]即可。
然后最后都减去他们自身即可满足三次。注意特判n = 1的情况
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<int,int> pii; const int N = 1e5+5; const int M = 1e6+5; const LL Mod = 199999; #define rg register #define pi acos(-1) #define INF 1e18 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < ‘0‘ || c > ‘9‘){if(c == ‘-‘) f = -1;c = getchar();} while(c >= ‘0‘ && c <= ‘9‘){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } void print(int x){ if(x < 0){x = -x;putchar(‘-‘);} if(x > 9) print(x/10); putchar(x%10+‘0‘); } } using namespace FASTIO; void FRE(){ /*freopen("data1.in","r",stdin); freopen("data1.out","w",stdout);*/} LL b[N],a[N]; int main() { int n;n = read(); for(rg int i = 1;i <= n;++i) a[i] = read(); if(n == 1) { printf("1 1\n"); printf("1\n"); printf("1 1\n"); printf("1\n"); printf("1 1\n"); printf("%d\n",-a[1]-2); } else { for(rg int i = 1;i < n;++i) b[i] = 1LL*(n-1)*a[i],a[i] += b[i]; printf("1 %d\n",n-1); for(rg int i = 1;i < n;++i) printf("%lld%c",b[i],i == n-1 ? ‘\n‘ : ‘ ‘); printf("%d %d\n",n,n); printf("%lld\n",1LL*n*a[n]-a[n]); a[n] = 1LL*n*a[n]; printf("%d %d\n",1,n); for(rg int i = 1;i <= n;++i) printf("%lld%c",-a[i],i == n ? ‘\n‘ : ‘ ‘); } // system("pause"); }
D：博弈论
暴力解法：肯定选更大的好，那么模拟一下选的过程即可
可以发现，当某个堆的数都大于其他堆的和时。
先手选这个堆必胜。
当不满足时，最后肯定会选到剩下1.那么判断下所有堆的和的奇偶性即可。
// Author: levil #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<int,int> pii; const int N = 1e5+5; const int M = 1e6+5; const LL Mod = 199999; #define rg register #define pi acos(-1) #define INF 1e18 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < ‘0‘ || c > ‘9‘){if(c == ‘-‘) f = -1;c = getchar();} while(c >= ‘0‘ && c <= ‘9‘){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } void print(int x){ if(x < 0){x = -x;putchar(‘-‘);} if(x > 9) print(x/10); putchar(x%10+‘0‘); } } using namespace FASTIO; void FRE(){ /*freopen("data1.in","r",stdin); freopen("data1.out","w",stdout);*/} int a[105]; int main() { int ca;ca = read(); while(ca--) { int mx = -1,sum = 0; int n;n = read(); for(rg int i = 1;i <= n;++i) a[i] = read(),sum += a[i],mx = max(mx,a[i]); if(mx > sum-mx) printf("T\n"); else { if(sum&1) printf("T\n"); else printf("HL\n"); } } // system("pause"); }
《Codeforces Round #666 (Div. 2)》
原文：https://www.cnblogs.com/zwjzwj/p/13591716.html
