计蒜客信息学5月入门赛;
题目要求不再叙述了;
A:
注意:数据类型 long long
#include<iostream>
#include<cstdio>
using namespace std;
#define ll long long
const int maxa=1e3+10;
ll a,b,c,d;
int main(){
scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
printf("%lld\n",a*d+b*c);
return 0;
}
B:
#include<iostream>
#include<cstdio>
using namespace std;
#define ll long long
const int maxa=1e3+10;
int main(){
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
for(int j=1;j<=2*n-i;j++){
if(j<=n-i) printf(" ");
else printf("A");
if(j==2*n-i) printf("\n");
}
return 0;
}
C:
#include<iostream>
#include<cstdio>
using namespace std;
#define ll long long
const int maxa=1e3+10;
int main(){
ll ans=0;
string s;
cin>>s;
for(int i=0;i<s.size();i++)
if(s[i]>='a'&&s[i]<='z') ans+=s[i]-'a'+1;
else if(s[i]>='A'&&s[i]<='Z') ans+=s[i]-'A'+1;
printf("%lld\n",ans);
return 0;
}
D:
思路:数组中快速查找两个数之和为k,方法很多,这里我提供了O(n)时间复杂度的算法,比较快;
其余方法可见链接:https://blog.csdn.net/queque_heiya/article/details/106155584
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
const int maxa=1e3+10;
ll n,k,ans=0;
ll a[maxa];
bool cnt(){
sort(a,a+n);
for(int i=0,j=n-1;i<j;){
if(a[i]+a[j]==k) return true;
else if(a[i]+a[j]<k) i++;
else j--;
}
return false;
}
int main(){
scanf("%lld%lld",&n,&k);
for(int i=0;i<n;i++){
for(int j=0;j<n;j++)
scanf("%lld",&a[j]);
ans+=(int)cnt();
}
printf("%lld\n",ans);
}