3 条题解
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1zxj (zhangxuanju) LV 7 @ 2023-11-30 12:52:40
#include <bits/stdc++.h> using namespace std; #define LL long long const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; int a[1005], b[1005], c[1005]; char s1[1005], s2[1005]; int main() { int n; cin >> n >> s1 >> s2; int l = strlen(s1); for (int i = 0; i < l; i++) if (s1[i] >= 'a' && s1[i] <= 'z') a[l - i] = s1[i] - 87; else a[l - i] = s1[i] - 48; l = strlen(s2); for (int i = 0; i < l; i++) if (s2[i] >= 'a' && s2[i] <= 'z') b[l - i] = s2[i] - 87; else b[l - i] = s2[i] - 48; l = max(strlen(s1), strlen(s2)); for (int i = 1; i <= l; i++) { c[i] += a[i] + b[i]; c[i + 1] = c[i] / n; c[i] %= n; } if (c[l + 1]) l++; for (int i = l; i > 0; i--) if (c[i] > 9) cout << char(c[i] + 87); else cout << c[i]; return 0; }
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02023-12-6 13:32:48@
这题只要注意9以上的数字的处理就行了。
上AC代码
#include<bits/stdc++.h> using namespace std; const int N=1e3+10; long long n,a[N],b[N],c[N]; int main(){ #define int long long char s1[N],s2[N]; cin>>n>>s1>>s2; int l=strlen(s1),r=strlen(s2);int s=max(l,r); for(int i=0;i<l;i++){ if(s1[i]>='a'&&s1[i]<='z')a[l-i]=s1[i]-87; else a[l-i]=s1[i]-48; } for(int i=0;i<r;i++){ if(s2[i]>='a'&&s2[i]<='z')b[r-i]=s2[i]-87; else b[r-i]=s2[i]-48; } for(int i=0;i<=s;i++){ c[i]+=a[i]+b[i]; c[i+1]+=c[i]/n; c[i]%=n; } if(c[s+1]){ s++; } for(int i=s;i>0;i--){ if(c[i]>9)cout<<char(c[i]+87); else cout<<c[i]; } return 0; }
点个赞吧,球球了。
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02022-1-8 16:41:00@
/***************************************** 备注: ******************************************/ #include <queue> #include <math.h> #include <stack> #include <stdio.h> #include <iostream> #include <vector> #include <iomanip> #include <string.h> #include <algorithm> using namespace std; #define LL long long const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; int a[1005], b[1005], c[1005]; char s1[1005], s2[1005]; int main() { int n; cin >> n >> s1 >> s2; int l = strlen(s1); for (int i = 0; i < l; i++) if (s1[i] >= 'a' && s1[i] <= 'z') a[l - i] = s1[i] - 87; else a[l - i] = s1[i] - 48; l = strlen(s2); for (int i = 0; i < l; i++) if (s2[i] >= 'a' && s2[i] <= 'z') b[l - i] = s2[i] - 87; else b[l - i] = s2[i] - 48; l = max(strlen(s1), strlen(s2)); for (int i = 1; i <= l; i++) { c[i] += a[i] + b[i]; c[i + 1] = c[i] / n; c[i] %= n; } if (c[l + 1]) l++; for (int i = l; i > 0; i--) if (c[i] > 9) cout << char(c[i] + 87); else cout << c[i]; return 0; }
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信息
- ID
- 1193
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 5
- 标签
- 递交数
- 118
- 已通过
- 46
- 上传者