2020.10.09 Rating 補題報告
阿新 • • 發佈：2020-10-18
A - A Blend of Springtime
題意：找給出的字串是否包含連續的不同的且不為“.”的三個字元，
做法：emmmm就是從頭讀之後判斷就行
程式碼：
//去吧馬里奧！把AC公主救回來！ // ******** // ************ // ####....#. // #..###.....##.... // ###.......###### // ........... // ##*####### // ####*******###### // ...#***.****.*###.... // ....**********##.....// ....**** *****.... // #### #### // ###### ###### #include<iostream> #include<algorithm> #include<cstdio> #include<cmath> #include<string> #include<map> #include<sstream> #include<cstring> #include<vector> #include<iomanip> #include<queue> #include<set> #define LL long long #define _64 __int64 const double PI = atan(1.)*4.; using namespace std; char a[2000]; int main(){ cin >> a; int flag = 0; //cout << a << endl; //cout << sizeof(a) << endl; for(int i = 1;i < strlen(a)-1;i++){ if(a[i] != a[i+1] && a[i] != a[i-1] && a[i+1] != a[i-1] && a[i] != '.' && a[i+1] != '.' && a[i-1] != '.'){ flag = 1; break; } } if(flag == 1){ cout << "Yes" << endl; }else{ cout << "No" << endl; } }
B - A Tide of Riverscape
做法：判斷，如果該點是“.”，直接全部變成0，之後判斷它相對的是不是1，用一個flag記錄即可
程式碼：
//去吧馬里奧！把AC公主救回來！ // ******** // ************ // ####....#. // #..###.....##.... // ###.......###### // ........... // ##*####### // ####*******###### // ...#***.****.*###.... // ....**********##..... // ....**** *****.... // #### #### // ###### ###### #include<iostream> #include<algorithm> #include<cstdio> #include<cmath> #include<string> #include<map> #include<sstream> #include<cstring> #include<vector> #include<iomanip> #include<queue> #include<set> #define LL long long #define _64 __int64 const double PI = atan(1.)*4.; using namespace std; int main(){ int n,p; cin >> n >> p; string s; cin >> s; int flag = 0; for(int i = 0;i < n;i++){ if(s[i] == '.'){ s[i] = '0'; if(i + p < n && s[i+p] == '0' || i - p >= 0 && s[i-p] == '0'){ s[i] = '1'; } } if(i + p < n && s[i] != s[i+p]){ flag = 1; } } if(flag == 1){ cout << s << endl; }else{ cout << "No" << endl; } }
D - Infinity Gauntlet
題意：一組資料有一個顏色還有一個名字，其中有對應關係，給出顏色輸出名字
做法：用陣列存了（其實感覺用map會簡單些），再輸出就行
程式碼：
//去吧馬里奧！把AC公主救回來！ // ******** // ************ // ####....#. // #..###.....##.... // ###.......###### // ........... // ##*####### // ####*******###### // ...#***.****.*###.... // ....**********##..... // ....**** *****.... // #### #### // ###### ###### #include<iostream> #include<algorithm> #include<cstdio> #include<cmath> #include<string> #include<map> #include<sstream> #include<cstring> #include<vector> #include<iomanip> #include<queue> #include<set> #define LL long long #define _64 __int64 const double PI = atan(1.)*4.; using namespace std; string a[10]; string b[10]; int mark[10]; int main(){ int n; cin >> n; for(int i = 0;i <10;i++)mark[i] = 0; a[1] = "Power";b[1] = "purple"; a[2] = "Time";b[2] = "green"; a[3] = "Space";b[3] = "blue"; a[4] = "Soul";b[4] = "orange"; a[5] = "Reality";b[5] = "red"; a[6] = "Mind";b[6] = "yellow"; if(n == 0){ for(int i=1;i < 10;i++){ mark[i] = 0; } }else{ for(int i = 0;i < n;i++){ string s; cin >> s; if(s == "purple"){ mark[1] =1; }else if(s == "green"){ mark[2] = 1; }else if(s == "blue"){ mark[3] = 1; }else if(s == "orange"){ mark[4] = 1; }else if(s == "red"){ mark[5] = 1; }else if(s == "yellow"){ mark[6] = 1; } } } cout << 6-n << endl; //cout << mark[3] << endl; for(int i = 1;i < 7;i++){ if(mark[i] == 0){ cout << a[i] << endl; } } }
E - High School: Become Human
題意：都明白
做法：取對數，e的ln次方，一通亂算可以得出這個式子就等於x的y次方，可以通過比較對數的方法比較這倆數，另外由於是double型別有精度問題，也就是比較是否相等的時候得通過差多麼接近0得出，這個精度得取得很小才行
程式碼：
//去吧馬里奧！把AC公主救回來！ // ******** // ************ // ####....#. // #..###.....##.... // ###.......###### // ........... // ##*####### // ####*******###### // ...#***.****.*###.... // ....**********##..... // ....**** *****.... // #### #### // ###### ###### #include<iostream> #include<algorithm> #include<cstdio> #include<cmath> #include<string> #include<map> #include<sstream> #include<cstring> #include<vector> #include<iomanip> #include<queue> #include<set> #define LL long long #define _64 __int64 const double PI = atan(1.)*4.; using namespace std; int main(){ double x,y; cin >> x >> y ; double ansx,ansy; //LL a,b; //a = x/10; //b = y/10; //ansx = pow(a,b); //ansy = pow(b,a); //cout << ansx << ansy << endl; //ansx = log(x)/log(y); //ansy = log(y)/log(x); ansx = y*log(x); ansy = x*log(y); //if(ansx > ansy){ // cout << '>' << endl; //}else if(ansx < ansy){ // cout << '<' << endl; //}else if(fabs(ansx - ansy) < 1e-16){ //cout << '=' << endl; //} if(fabs(ansx - ansy) <= 1e-100){ cout << '=' << endl; }else if(ansx - ansy < 0){ cout << '<' << endl; }else{ cout << '>' << endl; } }
F - Three displays
題意：找既滿足i<j<k，且滿足si<sj<sk的最少的租金
程式碼：
//去吧馬里奧！把AC公主救回來！ // ******** // ************ // ####....#. // #..###.....##.... // ###.......###### // ........... // ##*####### // ####*******###### // ...#***.****.*###.... // ....**********##..... // ....**** *****.... // #### #### // ###### ###### #include<iostream> #include<algorithm> #include<cstdio> #include<cmath> #include<string> #include<map> #include<sstream> #include<cstring> #include<vector> #include<iomanip> #include<queue> #include<set> #define LL long long #define _64 __int64 const double PI = atan(1.)*4.; using namespace std; struct ans{ int s; LL rent; }anss[5000]; int main(){ int n; cin >> n; for(int i = 0;i < n;i++){ cin >> anss[i].s; } for(int i = 0;i < n;i++){ cin >> anss[i].rent; } int flag1 = 0; int flag2 = 0; int flag = 0; LL minnx = 1e18; for(int i = 1;i < n-1;i++){ LL ansx = anss[i].rent; flag1 = 0; flag2 = 0; LL minn1 = 1e8; LL minn2 = 1e8; for(int j = 0;j < i;j++){ if(anss[j].s < anss[i].s){ minn1 = min(minn1,anss[j].rent); flag1 = 1; } } for(int j = i+1;j < n;j++){ if(anss[j].s > anss[i].s){ minn2 = min(minn2,anss[j].rent); flag2 = 1; } } if(flag1 ==1 && flag2 == 1){ flag = 1; ansx += minn1; ansx += minn2; minnx = min(minnx,ansx); } } if(flag == 1){ cout << minnx << endl; }else{ cout << -1 << endl; } }