F DPS
(坑点) 如果使用int会爆,使用double会缺失精度,因此使用long long。
代码
#include<bits/stdc++.h> using namespace std; typedef long long ll; ll a[110]; int main() { memset(a,0,sizeof(a)); ll n,i,ma=-1; scanf("%lld",&n); for(i=0; i<n; i++) { scanf("%lld",&a[i]); ma=max(ma,a[i]); } for(i=0; i<n; i++) { printf("+"); ll s=ceil(50*a[i]*1.0/ma); for(ll j=0; j<s; j++) { printf("-"); } printf("+\n"); printf("|"); for(ll j=0; j<s-1; j++) { printf(" "); } if(a[i]==ma) { printf("*"); } else if(a[i]!=0) { printf(" "); } printf("|"); printf("%d\n",a[i]); printf("+"); for(ll j=0; j<s; j++) { printf("-"); } printf("+\n"); } }
来自pdf
题意:
• 有一个无穷大的二维网格,每个格子可以是1、2或者3,每个1旁边要有一个2和3,要使机器的占比 最大。
做法:
• (i+j)%3 =0 交错2和3
• 答案是2/3 • 因为这样每个1旁边恰好有一个2和3,而任意两个2和3不相邻。
• 可以计算出这是上限。
代码略
关键在于,大数板子(Java和python选手除外)和如何找环
大数板子
int max(int a, int b) { return a>b?a:b; } struct bign { int len, s[numlen]; bign() { memset(s, 0, sizeof(s)); len = 1; } bign(int num) { *this = num; } bign(const char *num) { *this = num; } bign operator = (const int num) { char s[numlen]; sprintf(s, "%d", num); *this = s; return *this; } bign operator = (const char *num) { len = strlen(num); while(len > 1 && num[0] == '0') num++, len--; for(int i = 0;i < len; i++) s[i] = num[len-i-1] - '0'; return *this; } void deal() { while(len > 1 && !s[len-1]) len--; } bign operator + (const bign &a) const { bign ret; ret.len = 0; int top = max(len, a.len) , add = 0; for(int i = 0;add || i < top; i++) { int now = add; if(i < len) now += s[i]; if(i < a.len) now += a.s[i]; ret.s[ret.len++] = now%10; add = now/10; } return ret; } bign operator - (const bign &a) const { bign ret; ret.len = 0; int cal = 0; for(int i = 0;i < len; i++) { int now = s[i] - cal; if(i < a.len) now -= a.s[i]; if(now >= 0) cal = 0; else { cal = 1; now += 10; } ret.s[ret.len++] = now; } ret.deal(); return ret; } bign operator * (const bign &a) const { bign ret; ret.len = len + a.len; for(int i = 0;i < len; i++) { for(int j = 0;j < a.len; j++) ret.s[i+j] += s[i]*a.s[j]; } for(int i = 0;i < ret.len; i++) { ret.s[i+1] += ret.s[i]/10; ret.s[i] %= 10; } ret.deal(); return ret; } //乘以小数,直接乘快点 bign operator * (const int num) { bign ret; ret.len = 0; int bb = 0; for(int i = 0;i < len; i++) { int now = bb + s[i]*num; ret.s[ret.len++] = now%10; bb = now/10; } while(bb) { ret.s[ret.len++] = bb % 10; bb /= 10; } ret.deal(); return ret; } bign operator / (const bign &a) const { bign ret, cur = 0; ret.len = len; for(int i = len-1;i >= 0; i--) { cur = cur*10; cur.s[0] = s[i]; while(cur >= a) { cur -= a; ret.s[i]++; } } ret.deal(); return ret; } bign operator % (const bign &a) const { bign b = *this / a; return *this - b*a; } bign operator += (const bign &a) { *this = *this + a; return *this; } bign operator -= (const bign &a) { *this = *this - a; return *this; } bign operator *= (const bign &a) { *this = *this * a; return *this; } bign operator /= (const bign &a) { *this = *this / a; return *this; } bign operator %= (const bign &a) { *this = *this % a; return *this; } bool operator < (const bign &a) const { if(len != a.len) return len < a.len; for(int i = len-1;i >= 0; i--) if(s[i] != a.s[i]) return s[i] < a.s[i]; return false; } bool operator > (const bign &a) const { return a < *this; } bool operator <= (const bign &a) const { return !(*this > a); } bool operator >= (const bign &a) const { return !(*this < a); } bool operator == (const bign &a) const { return !(*this > a || *this < a); } bool operator != (const bign &a) const { return *this > a || *this < a; } string str() const { string ret = ""; for(int i = 0;i < len; i++) ret = char(s[i] + '0') + ret; return ret; } }; 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; } // 大数开平方 bign Sqrt(bign x) { int a[numlen/2]; int top = 0; for(int i = 0;i < x.len; i += 2) { if(i == x.len-1) { a[top++] = x.s[i]; } else a[top++] = x.s[i] + x.s[i+1]*10; } bign ret = (int)sqrt((double)a[top-1]); int xx = (int)sqrt((double)a[top-1]); bign pre = a[top-1] - xx*xx; bign cc; for(int i = top-2;i >= 0; i--) { pre = pre*100 + a[i]; cc = ret*20; for(int j = 9;j >= 0; j--) { bign now = (cc + j)*j; if(now <= pre) { ret = ret*10 + j; pre -= now; break; } } } return ret; }
找环
for(int i=1; i<=n; i++) { if(!vis[i]) { num.clear(); int tmp=i; while(!vis[tmp]) { vis[tmp]=1; num.push_back(tmp); tmp=a[tmp]; } f(); } }
vis数组用来存放每个数是否被查询的状态,当未被查询时,进入循环,将当前这个数push_back到num容器中去。i从1到n为止。