A. Donut Shops
- 题目传送门
- 水题，就用a * b和c比一下大小就行了，要求买第一种饼干便宜的情况就假设只买一块，比较a和c的大小，求买第二种便宜就比较 a * b和c比大小就行了
#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize(2)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
ll t,a,b,c;
int main()
{
ios::sync_with_stdio(false);
cin>>t;
while(t--){
cin>>a>>b>>c;
ll a1=0,a2=0;
if(c>a){
cout<<"1"<<" ";
}else{
cout<<"-1"<<" ";
}
ll tmp=a*b;
if(tmp>c){
cout<<b<<"\n";
}else{
cout<<"-1"<<"\n";
}
}
return 0;
}
B. 01 Game
- 题目传送门
- 之前做过一个类似的，结果被自己误导了wa了一发，想复杂了，其实就记录一下0和1的个数然后看较小者能不能被2整除就行了，emmmm其实可以证明，1和0数量少的那一个总会被消完。
#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize(2)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
int t;
string s;
int main()
{
ios::sync_with_stdio(false);
cin>>t;
while(t--){
cin>>s;
int ans=0;
int a0=0,a1=0;
for(int i=0;i<s.size();++i){
if(s[i]=='1') ++a1;
else ++a0;
}
ans=min(a0,a1);
if(ans&1) cout<<"DA"<<"\n";
else cout<<"NET"<<"\n";
}
return 0;
}
C. Pluses and Minuses
-
这个其实很简单就把cur=0开始，把每个符号的cur都走一遍，比如
-
--+- -
对应 -1 -2 -1 -2
对应的为cur数组（代码中就是r数组）
那么就从-1开始走，标记ans=0，如果遇到比cur[i]更小的，那么ans就加上这个cur的下标。最后再加一个整个数组的长度 -
这题不开long long 会爆int，wa on test7，我就wa了一发（彩笔落泪
#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize(2)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
ll t;
string s;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
ll r[maxn];
int main()
{
ios::sync_with_stdio(false);
cin >> t;
while (t--) {
cin >> s;
//int rec=0;
//int a=0,r=0;
if (s[0] == '+') r[0] = 1;
else r[0] = -1;
ll len = s.size();
for (ll i = 1; i < s.size(); ++i) {
r[i] = s[i] == '+' ? r[i - 1] + 1 : r[i - 1] - 1;
}
//r[len] = -0x3f3f3f3f;
ll rec = 0;
ll ans = 0;
for (ll i = 0; i <= len; ++i) {
if (i == len) {
ans += len;
break;
}
if (r[i] < rec) {
rec = r[i];
ans += i + 1;
}
}
cout << ans << "\n";
}
return 0;
}
D. Maximum Sum on Even Positions
-
这个题比赛的时候想假了，最后还是看的dalao的（pxy，yyds！）下来之后其实仔细想想，这题是以前做过的一道题，只是这里把奇偶分开考虑了而已，我真的太菜了，这题其实只要发现只有偶数长度的子串翻转才有效，奇数长度的反转对结果没影响，然后再简单想想就有了（我真的太菜了，落泪
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
typedef long long ll;
ll t, n, a[maxn], r1[maxn], r2[maxn];
int main()
{
ios::sync_with_stdio(false);
cin >> t;
while (t--) {
cin >> n;
for (int i = 0; i < n; ++i) cin >> a[i];
int f = 1;
for (int i = 0; i < n - 1; i += 2) {
r1[f] = a[i + 1] - a[i];
f++;
}
int f1 = 1;
for (int i = 2; i < n; i += 2) {
r2[f1] = a[i - 1] - a[i];
f1++;
}
ll sum = 0;
for (int i = 0; i < n; i += 2) sum += a[i];
ll tmp = 0, now = 0;
for (int i = 0; i < f; ++i) {
now += r1[i];
if (now <= 0) now = 0;
tmp = max(tmp, now);
}
ll tmp1 = 0; now = 0;
for (int i = 0; i < f1; ++i) {
now += r2[i];
if (now <= 0) now = 0;
tmp1 = max(tmp1, now);
}
sum = max(sum + tmp1, sum + tmp);
cout << sum << "\n";
}
return 0;
}
顺便把大佬的代码也放上来叭（pxy，yyds！
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int s[200010],d[200010];
long long MaxSubseqSum4( int A[], int N ) {
long long ThisSum, MaxSum, i;
ThisSum = MaxSum = 0;
for( i = 1; i < N; i+=2 ) {
ThisSum += (A[i]);
if( ThisSum > MaxSum ){
MaxSum = ThisSum;
}
else if( ThisSum < 0 )
ThisSum = 0;
}
ThisSum=0;
for( i = 2; i < N; i+=2 ) {
ThisSum -= (A[i]);
if( ThisSum > MaxSum )
MaxSum = ThisSum;
else if( ThisSum < 0 )
ThisSum = 0;
}
return MaxSum;
}
int main(){
int a,b,c,t,x,y;
long long rzt,xxt,max;
scanf("%d",&t);
while(t--){
scanf("%d",&a);
rzt=0;
for(int l1=0;l1<a;l1++){
scanf("%d",s+l1);
if(!(l1%2))rzt+=s[l1];
}
for(int l1=1;l1<a;l1++){
d[l1]=s[l1]-s[l1-1];
}
rzt+=MaxSubseqSum4(d,a);
printf("%lld\n",rzt);
}
}
- 太菜了我，被自己菜哭，菜到睡不着，所以起来顺手把前四题的题解写了，ef明天睡醒了起来补