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100 Accepted A1000 【Example】Hello, World! Grammar 99 / 340 Beginner 100 Accepted A1001 【Example】A + B Problem Grammar 92 / 177 Beginner 100 Accepted A1002 Area of a Square 91 / 181 Beginner 100 Accepted A1003 Movie Tickets Grammar Basic Data Types 88 / 200 Beginner 100 Accepted A1004 Area of a Rectangle 89 / 115 Beginner 100 Accepted A1005 Perimeter of a Rectangle 88 / 122 Beginner 100 Accepted A1006 【Example】Division with Remainder 86 / 128 Beginner 100 Accepted A1007 Breaking Down Numbers and Summing Them 86 / 136 Beginner 100 Accepted A1008 Reverse Output a Three-Digit Number Breaking Down Numbers 85 / 197 Beginner 100 Accepted A1009 Reverse Output a Four-Digit Integer Breaking Down Numbers 78 / 245 Beginner 100 Accepted A1010 [ABC222A] Four Digits (Four Digits) 40 / 54 Beginner 100 Accepted A1011 【Example】Swapping Values Water Pouring Problem 81 / 197 Beginner 100 Accepted A1012 Cows Eating Grass Elementary Math Olympiad 72 / 119 Popularization- 100 Accepted A1013 Cuboid 78 / 250 Beginner 100 Accepted A1015 【Example】Floating Point Number with 3 Decimal Places 78 / 158 Beginner 100 Accepted A1016 Calculating the Floating Point Value of a Fraction 78 / 167 Beginner 100 Accepted A1017 Estimation of Earth's Population Carrying Capacity 57 / 152 Popularization- 100 Accepted A1018 Calculations Related to Circles 74 / 263 Beginner 100 Accepted A1019 Temperature Expression Conversion 73 / 197 Beginner 100 Accepted A1020 [ABC226A] Rounding Decimals (Round decimals) 41 / 54 Beginner 100 Accepted A1021 【Example】Size of Integer Data Type Storage Space 73 / 144 Beginner 100 Accepted A1022 Rounding Floating Point Numbers Towards Zero 71 / 136 Beginner 100 Accepted A1024 Print ASCII Code 75 / 159 Beginner 100 Accepted A1025 Print Character 71 / 134 Beginner 100 Accepted A1026 Size of Floating Point Data Type Storage Space -
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#include<bits/stdc++.h> using namespace std; struct tab{ char o; int x; }; tab bfe[]= { {'1', 13},{'2', 12}, {'3', 46}, {'4', 31}, {'5', 43}, {'6', 18},{'7', 40}, {'8', 28}, {'9', 5}, {'A', 54}, {'B', 20},{'C', 15}, {'D', 8}, {'E', 39}, {'F', 57}, {'G', 45},{'H', 36}, {'J', 38}, {'K', 51}, {'L', 42}, {'M', 49},{'N', 52}, {'P', 53}, {'Q', 7}, {'R', 4}, {'S', 9},{'T', 50}, {'U', 10}, {'V', 44}, {'W', 34}, {'X', 6},{'Y', 25}, {'Z', 1}, {'a', 26}, {'b', 29}, {'c', 56},{'d', 3}, {'e', 24}, {'f', 0}, {'g', 47}, {'h', 27},{'i', 22}, {'j', 41}, {'k', 16}, {'m', 11}, {'n', 37},{'o', 2}, {'p', 35}, {'q', 21}, {'r', 17}, {'s', 33},{'t', 30}, {'u', 48}, {'v', 23}, {'w', 55}, {'x', 32},{'y', 14}, {'z', 19} }; long long a[114]; int b[]={1,0,0,6,1,8,3,4,2,1,3,6,6,9,6,3,2,0}; int c[31],add[31]; int main() { int ad=0; string bv; cin>>bv; for(int i=0,l=1;i<10;i++,l++){ for(int j=0;j<60;j++){ if(bv[i]==bfe[j].o){ a[l]=bfe[j].x; break; } } } a[1]*=38068692544; a[2]*=3364; a[3]*=11316496; a[4]*=128063081718016; a[5]*=656356768; a[6]*=7427658739644928; a[7]*=195112; a[8]*=2207984167552; a[9]*=58; for(int i=1;i<=10;i++) ad+=a[i]; int la=0; reverse(b,b+18); for(int i=0;i<31;i++){ int t=ad; while(t){ t/=10; la++; } } for(int i=la;i>0;i++){ int t=ad; while(t){ add[i]=t%10; t/=10; } } int t=0; int lc=la; for(int i=0;i<lc;i++){ c[i]=add[i]-b[i]+t; if(c[i]<0){ c[i]+=10; t=-1; } else t=0; } while(lc>0&&c[lc]==0) lc--; ad=0; for(int i=0;i<lc;i++){ int j=i+1; long long x=1; for(int i=0;i<lc;i++) x*=10; ad+=c[i]*x; } int nor=ad^177451812; cout<<nor; }-
jike1994
@ 2025-5-15 16:21:00
#include<bits/stdc++.h> using namespace std; typedef long long LL; LL a[15]; int main() { map<char,LL> h={ {'1', 13},{'2', 12}, {'3', 46}, {'4', 31}, {'5', 43}, {'6', 18},{'7', 40}, {'8', 28}, {'9', 5}, {'A', 54}, {'B', 20},{'C', 15}, {'D', 8}, {'E', 39}, {'F', 57}, {'G', 45},{'H', 36}, {'J', 38}, {'K', 51}, {'L', 42}, {'M', 49},{'N', 52}, {'P', 53}, {'Q', 7}, {'R', 4}, {'S', 9},{'T', 50}, {'U', 10}, {'V', 44}, {'W', 34}, {'X', 6},{'Y', 25}, {'Z', 1}, {'a', 26}, {'b', 29}, {'c', 56},{'d', 3}, {'e', 24}, {'f', 0}, {'g', 47}, {'h', 27},{'i', 22}, {'j', 41}, {'k', 16}, {'m', 11}, {'n', 37},{'o', 2}, {'p', 35}, {'q', 21}, {'r', 17}, {'s', 33},{'t', 30}, {'u', 48}, {'v', 23}, {'w', 55}, {'x', 32},{'y', 14}, {'z', 19}}; a[0]=1; for(int i=1;i<=10;i++) { a[i]=a[i-1]*58; } string s; cin>>s; s.erase(0,2); LL sum=h[s[0]]*a[6]+h[s[1]]*a[2]+h[s[2]]*a[4]; sum+=h[s[3]]*a[8]+h[s[4]]*a[5]+h[s[5]]*a[9]+h[s[6]]*a[3]; sum+=h[s[7]]*a[7]+h[s[8]]*a[1]+h[s[9]]; LL b=100618342136696320; sum-=b; LL x=0b1010100100111011001100100100; sum=sum^x; cout<<sum; return 0; }
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#include<bits/stdc++.h> using namespace std; const int N=1e5+10; int p[N]; char s[N]; int n,m; bool check2(int x) { if((p[1]-1)/2>=x) return true; //头一个位置可以放置 if((n-p[m])/2>=x) return true; //后面一个位置可以放置 for(int i=1;i<m;i++) //区间可以放置两个 { int t=p[i+1]-p[i]; if(t/3>=x) return true; } return false; //找不到区间可以放置 } bool check3(int x) { int cnt=0; if(p[1]-1>=x) cnt++; if(n-p[m]>=x) cnt++; //if(x==3) cout<<cnt<<endl; for(int i=1;i<m;i++) { if((p[i+1]-p[i])/2>=x) cnt++; } if(cnt>=2) return true; return false; } int main() { cin>>n; cin>>s+1; for(int i=1;i<=n;i++) { if(s[i]=='1') { p[++m]=i; } } if(m==0) { cout<<n-1; } else { int l=0,r=n; for(int i=1;i<m;i++) //找到一个最小的区间 { l=min(l,p[i+1]-p[i]); } while(l<r) { int mid=(l+r+1)>>1; //cout<<mid<<" "<<check2(mid)<<" "<<check3(mid)<<endl; if(check2(mid)||check3(mid)) { l=mid; } else { r=mid-1; } } cout<<l; } return 0; } -
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#include<cstdio> #include<cstring> #include<vector> #include<set> using namespace std; struct sd { int a,b; bool operator < (const sd& c) const { if (a<c.a) ; return 1; if (a==c.a && b<c.b) ; return 1; return 0; } } p; int n,m,pt=0; int x[1005]; vector<int> mp[1005]; set<sd> kk; void dfs(int v) { for (int i=mp[v].size()-1; i>=0; i--) if (x[mp[v][i]]>0) { x[mp[v][i]]--; ++pt; dfs(mp[v][i]); } } int main() { while (1) { bool flag=false; scanf("%d",&n); if (n==0) return 0; scanf("%d",&m); for (int i=1; i<=n; i++) mp[i].clear(); memset(x,0,sizeof x); kk.clear(); for (int a,b,t,i=1; i<=m; i++) { scanf("%d%d",&a,&b); if (a==b) continue ; if (a>b) swap(a,b); p.a=a,p.b=b; if (kk.count(p)) continue; kk.insert(p); mp[a].push_back(b); ++x[a]; mp[b].push_back(a); ++x[b]; } for (int i=1; i<=n; i++) if (x[i]==0 || x[i] % 2==1) { flag=1; break; } if (flag) { printf("0\n"); continue; } x[1]=-1; pt=1; dfs(1); // if (pt==n) printf("1\n"); } return 0; }
熊梓诺
LV 3
@ 2025-6-7 11:34:27
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