1 条题解
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0
C :
#include<stdio.h> int main() { int n,i,sum,big; sum=0; big=0; scanf("%d",&n); int a[n],b[n],c[n]; for(i=0;i<n;i++) scanf("%d",&a[i]); for(i=0;i<n;i++) scanf("%d",&b[i]); for(i=0;i<n;i++) { if(a[i]>=b[i]) sum+=a[i]; else sum+=b[i]; } for(i=0;i<n;i++) { if(a[i]>=b[i]) c[i]=b[i]; else c[i]=a[i]; if(c[i]>big) big=c[i]; } sum+=big; printf("%d",sum); }C++ :
#include <bits/stdc++.h> #define int long long using namespace std; const int N=1e6+10; typedef pair<int,int> PII; typedef long long ll; typedef priority_queue<int , vector<int>, greater<int>> minqueue; //′óD?μ?′ó queue typedef priority_queue<int, vector<int>, less<int>> maxqueue; //′ó′óμ?D? queue int gcd(int a, int b){ //×?′ó1?òòêy return b ? gcd(b, a % b) : a; } int lcm(int a, int b){ //×?D?1?±?êy return a * b / __gcd(a, b); } int qmi(int base, int power, int p) //?ì?ù?Y?óóà { int result = 1; while (power > 0) { if (power & 1) result = result * base % p; base = base * base % p ; power >>= 1; } return result % p; } int a[5][100010]; void solve() { int n; cin>>n; for(int i=1;i<=2;i++){ for(int j=1;j<=n;j++){ cin>>a[i][j]; } } int mx=-1e9; int sum1=0,sum2=0; int k=0; int mn=-1e9; for(int i=1;i<=n;i++){ mn=max(mn,min(a[1][i],a[2][i])); } int ans=0; for(int i=1;i<=n;i++){ ans+=max(a[1][i],a[2][i]); } ans+=mn; cout<<ans<<'\n'; } signed main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); int t; t=1; //cin>>t; while(t--) { solve(); } return 0; }
- 1
信息
- ID
- 738
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 10
- 标签
- 提交数
- 12
- 已通过
- 2
- 上传者