10 solutions
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0
#include<bits/stdc++.h> using namespace std; string s[50]={"glass","leaf","corn","rice","light","triangle","yucca","bone","relic","plank","basil","soil","shovel","bgg","pincer","privet","iris","powder","cactus","bubble","air","starfish","sand","sponge","lightning","claw","root","dahlia","bur","missiel","stinger","orange","wax","honey","poo","wing","faster","talisman","dice","chip","battery","sawblade","fragment","totem","mimic","corruption","crown","moon","yggdrasil","compass"},di[5]={"","legendary","mythic","ultra","super"}; map<string,int> cnt[5],p,id; int random(int n) { return 1ll*rand()*rand()%n; } int main() { srand((unsigned)time(0)); cout<<"you have these legendary petal:"<<endl; for (int i=0;i<50;i++) cnt[1][s[i]]=100000/可修改,不能超过1000000/,cout<<s[i]<<" "<<cnt[1][s[i]]<<endl; id["legendary"]=1; id["mythic"]=2; id["ultra"]=3; id["super"]=4; cout<<"now start to craft!"<<endl; while(true)//输入hxxxxxh结束合卡 { string lev,pt; int x,sum=0; cin>>lev; if(lev=="hxxxxxh") break; cin>>pt>>x; if(cnt[id[lev]][pt]<x) { cout<<"fuck you noob宸!"<<endl; continue; } x=x/5; for (int i=1;i<=x;i++) { int y=random(100); if(id[lev]1&&y%240) sum++; if(id[lev]2&&y%450) sum++; if(id[lev]3&&y50) sum++; } cnt[id[lev]+1][pt]+=sum; cnt[id[lev]][pt]=random(5); cout<<"you have crafted "<<sum<<" "<<di[id[lev]+1]<<" "<<pt<<"!!!!!"<<endl; cout<<"now you have these level "<<pt<<":"<<endl; for (int i=1;i<=4;i++) cout<<di[i]<<" "<<cnt[i][pt]<<endl; } bool flg=0; //legendary和mythic过于noob,所以只输出ultra和super花瓣 cout<<"ultra petal:"<<endl; for (int j=0;j<50;j++) if(cnt[3][s[j]]) cout<<s[j]<<" "<<cnt[3][s[j]]<<endl; cout<<"super petal:"<<endl; for (int j=0;j<50;j++) if(cnt[4][s[j]]) cout<<s[j]<<" "<<cnt[4][s[j]]<<endl,flg=1; //出super的是pro,没出的是noob if(flg) cout<<"pro!!!"<<endl; else cout<<"noob..."<<endl; return 0; }
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-4
很明显,新手可能并不知道如何使用加法符号,所以我们可以通过高精度来实现
#include<bits/stdc++.h> using namespace std; int a[250],b[250],c[250]; int main(){ string A,B; cin>>A>>B; int len=max(A.length(),B.length()); for(int i=A.length()-1,j=1;i>=0;i--,j++)a[j]=A[i]-'0'; for(int i=B.length()-1,j=1;i>=0;i--,j++)b[j]=B[i]-'0'; for(int i=1;i<=len;i++){ c[i]+=a[i]+b[i]; c[i+1]=c[i]/10; c[i]%=10; } if(c[len+1])len++; for(int i=len;i>=1;i--)cout<<c[i]; return 0; } //纯恶意,无娱乐
- 1
Information
- ID
- 6762
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- 1
- Tags
- # Submissions
- 406
- Accepted
- 90
- Uploaded By