牛客多校训练AFJ(签到)-白红宇

牛客多校训练AFJ(签到)

阅读量：4327 次

发布时间：2019-06-06

本文共 4889 字，大约阅读时间需要 16 分钟。

题目链接（单调栈）

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             #include
              
               const int maxn=1e5+5;typedef long long ll;using namespace std;int L1[5*maxn],L2[5*maxn];int a[5*maxn],b[5*maxn];int main(){ int n; while(~scanf("%d",&n)) { for(int t=1;t<=n;t++) { scanf("%d",&a[t]); } for(int t=1;t<=n;t++) { scanf("%d",&b[t]); } stack
               
                s1,s2; for(int t=1;t<=n;t++) { if(s1.empty()) { L1[t]=0; s1.push(t); continue; } else { while(!s1.empty()&&a[s1.top()]>a[t]) { s1.pop(); } if(s1.empty()) { L1[t]=0; s1.push(t); continue; } else { L1[t]=s1.top(); s1.push(t); continue; } } } for(int t=1;t<=n;t++) { if(s2.empty()) { L2[t]=0; s2.push(t); continue; } else { while(!s2.empty()&&b[s2.top()]>b[t]) { s2.pop(); } if(s2.empty()) { L2[t]=0; s2.push(t); continue; } else { L2[t]=s2.top(); s2.push(t); continue; } } } int k=n+1; for(int t=1;t<=n;t++) { //cout<
                
                 <<" "<
                 
                  <

题目链接：（思维）

三角形面积的22倍

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             #include
              
               const int maxn=1e5+5;typedef long long ll;using namespace std;int main(){ long double x0,y0,x1,y1,x2,y2; long double a,b,c,p,S; long double ans; while(cin>>x0>>y0>>x1>>y1>>x2>>y2) { a=sqrtl((x0-x1)*(x0-x1)+(y0-y1)*(y0-y1)); b=sqrtl((x0-x2)*(x0-x2)+(y0-y2)*(y0-y2)); c=sqrtl((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); //cout<
               <<" "<<<" "<
                 
                  <

海伦公式版没过不知道原因

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             #include
              
               const int maxn=1e5+5;typedef long long ll;using namespace std;int main(){ long double x0,y0,x1,y1,x2,y2; long double a,b,c,p,S; long double ans; while(cin>>x0>>y0>>x1>>y1>>x2>>y2) { a=sqrtl((x0-x1)*(x0-x1)+(y0-y1)*(y0-y1)); b=sqrtl((x0-x2)*(x0-x2)+(y0-y2)*(y0-y2)); c=sqrtl((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); //cout<
               <<" "<<<" "<
                 
                  <

题目链接：（大数）

#include 
     
      #include 
      
       #include 
       
         #include 
        
          #include 
         
           #include 
          
            using namespace std; const int maxn = 1000; struct bign{ int d[maxn], len; void clean() { while(len > 1 && !d[len-1]) len--; } bign() { memset(d, 0, sizeof(d)); len = 1; } bign(int num) { *this = num; } bign(char* num) { *this = num; } bign operator = (const char* num){ memset(d, 0, sizeof(d)); len = strlen(num); for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0'; clean(); return *this; } bign operator = (int num){ char s[20]; sprintf(s, "%d", num); *this = s; return *this; } bign operator + (const bign& b){ bign c = *this; int i; for (i = 0; i < b.len; i++){ c.d[i] += b.d[i]; if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++; } while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++; c.len = max(len, b.len); if (c.d[i] && c.len <= i) c.len = i+1; return c; } bign operator - (const bign& b){ bign c = *this; int i; for (i = 0; i < b.len; i++){ c.d[i] -= b.d[i]; if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--; } while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--; c.clean(); return c; } bign operator * (const bign& b)const{ int i, j; bign c; c.len = len + b.len; for(j = 0; j < b.len; j++) for(i = 0; i < len; i++) c.d[i+j] += d[i] * b.d[j]; for(i = 0; i < c.len-1; i++) c.d[i+1] += c.d[i]/10, c.d[i] %= 10; c.clean(); return c; } bign operator / (const bign& b){ int i, j; bign c = *this, a = 0; for (i = len - 1; i >= 0; i--) { a = a*10 + d[i]; for (j = 0; j < 10; j++) if (a < b*(j+1)) break; c.d[i] = j; a = a - b*j; } c.clean(); return c; } bign operator % (const bign& b){ int i, j; bign a = 0; for (i = len - 1; i >= 0; i--) { a = a*10 + d[i]; for (j = 0; j < 10; j++) if (a < b*(j+1)) break; a = a - b*j; } return a; } bign operator += (const bign& b){ *this = *this + b; return *this; } bool operator <(const bign& b) const{ if(len != b.len) return len < b.len; for(int i = len-1; i >= 0; i--) if(d[i] != b.d[i]) return d[i] < b.d[i]; return false; } bool operator >(const bign& b) const{ return b < *this;} bool operator<=(const bign& b) const{ return !(b < *this);} bool operator>=(const bign& b) const{ return !(*this < b);} bool operator!=(const bign& b) const{ return b < *this || *this < b;} bool operator==(const bign& b) const{ return !(b < *this) && !(b > *this);} string str() const{ char s[maxn]={}; for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0'; return s; } }; istream& operator >> (istream& in, bign& x) { string s; in >> s; x = s.c_str(); return in; } ostream& operator << (ostream& out, const bign& x) { out << x.str(); return out; }int main(){ bign x,y,a,b; while(cin>>x>>a>>y>>b) { if(x*b==y*a) { puts("="); } else if(x*b
           
            "); } }}