Comet OJ - Contest #8 A.B.C

题目

A.杀手皇后

签到题，直接用string进行比较即可

/*
Algorithm:
Author: anthony1314
Creat Time:
Time Complexity:
*/

#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cstring>
#include<cstdio>
//#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define mod 1e9 + 7
#define line printf("--------------");
using namespace std;

int main() {
	string maxx = "";
	maxx += string(1001, '~');
	int n;
	cin>>n;
	string q;
	for(int i = 0; i < n; i++) {
		cin>>q;
		if(q < maxx){
			maxx = q;
		}
	}
	cout<<maxx<<endl;
	return 0;
}

B.支援城市

如果有n=3, 三个数为a, b, c,则对于c

/*
Algorithm:
Author: anthony1314
Creat Time:
Time Complexity:
*/

#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cstring>
#include<cstdio>
//#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define mod 1e9 + 7
#define line printf("--------------");
using namespace std;
ll a[100005];
ll b[100005];
int main() {
	int n;
	cin>>n;
	ll sum1 = 0;
	ll sum2 = 0;
	for(int i = 0; i < n; i++){
		scanf("%lld", &a[i]);
		b[i] = a[i]*a[i];
		sum1 += b[i];
		sum2 += a[i];
	}
	for(int i = 0; i < n; i++){
		if(i != 0) printf(" ");
		printf("%lld", sum1+(n-2)*b[i]-2*a[i]*(sum2-a[i]));
	}
	puts("");


	return 0;
}

C.符文能量

n个符文石会爆发n-1次能量，且不影响，先算出n-1能是多少

开两个数组记录 k倍符文能量爆发是多少 k^2倍符文能量爆发是多少

直接dp就行了 dp三个状态还没区间在选区间已经选完区间之间爆发的最少能量

/*
Algorithm:
Author: anthony1314
Creat Time:
Time Complexity:
*/

#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cstring>
#include<cstdio>
//#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define mod 1e9 + 7
#define line printf("--------------");
using namespace std;
ll a[100005];
ll b[100005];
ll num[100005];
ll dp[100005][5];
ll knum[100005];
ll k2num[100005];
int main() {
	ll n, k;
	cin>>n>>k;
	for(int i = 0; i < n; i++){
		scanf("%lld %lld", &a[i], &b[i]);
	}
	for(int i = 1; i < n; i++){
		num[i] = a[i] * b[i-1];
		knum[i] = k*num[i];
		k2num[i] = k*k*num[i];
	}
	dp[1][0] = num[1];//没选 
	dp[1][1] = knum[1];//选完了 
	dp[1][2] = k2num[1];//还在选 
	for(int i = 2; i < n; i++){
		dp[i][0] = dp[i-1][0] + num[i];
		dp[i][1] = min(dp[i-1][1] + num[i], dp[i-1][2] + knum[i]);
		dp[i][2] = min(dp[i-1][0] + knum[i], dp[i-1][2] + k2num[i]);
	}
	ll minn = min(min(dp[n-1][0], dp[n-1][1]), dp[n-1][2]);
	cout<<minn<<endl;
	return 0;
}