组队
#include<stdio.h>
int main(void) {
freopen("team.txt","r",stdin);
int a[21][6];
for(int i=1; i<=20; ++i)
for(int j=0; j<=5; ++j)//每个人前面有一个序号
scanf("%d",&a[i][j]);
int ans=0;
for(int i1=1; i1<=20; ++i1) {
for(int i2=1; i2<=20; ++i2) {
if(i2==i1)continue;
for(int i3=1; i3<=20; i3++) {
if(i3==i1||i3==i2)continue;
for(int i4=1; i4<=20; ++i4) {
if(i4==i1||i4==i2||i4==i3)continue;
for(int i5=1; i5<=20; ++i5) {
if(i5==i1||i5==i2||i5==i3||i5==i4)continue;
if(a[i1][1]+a[i2][2]+a[i3][3]+a[i4][4]+a[i5][5]>ans)ans=a[i1][1]+a[i2][2]+a[i3][3]+a[i4][4]+a[i5][5];
}
}
}
}
}
printf("%d\n",ans);//490
return 0;
} #include<algorithm>
#include<cstdio>
#include<iostream>
using namespace std;
const int n=20;
const int m=5;
int a[n+1][m+1];
int vis[m+1];
int ans=0;
void dfs(int lv,int sum) {
if(lv==6) {
ans=max(ans,sum);
return;
}
for(int i=1; i<=n; ++i) {
bool f=0;
for(int j=1; j<lv; ++j) {
if(vis[j]==i) {
f=1;
}
}
if(!f) {
vis[lv]=i;
dfs(lv+1,sum+a[i][lv]);
}
}
}
int main() {
for(int i=1; i<=20; ++i)
for(int j=0; j<=5; ++j)//每个人前面有一个序号
scanf("%d",&a[i][j]);
dfs(1,0);
cout << ans << endl;
} 年号字串
#include<algorithm>
#include<iostream>
using namespace std;
int main() {
int a;
while(cin>>a) {
string ans="";
while(a) {
--a;//精髓
ans+='A'+a%26;
a/=26;
}
reverse(ans.begin(),ans.end());
cout<<ans<<endl;//BYQ
}
return 0;
} 数列求值
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
int a=1,b=1,c=1,d;
int main() {
for(int i=4;i<=20190324;++i){
d=(a+b+c)%10000;
a=b,b=c,c=d;
}cout<<d<<endl;//4659
return 0;
} 数的分解
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
bool chk(int n) {
for (; n; n /= 10)
if (n % 10 == 2 || n % 10 == 4) return 0;
return 1;
}
int main() {
int cnt = 0;
for (int i = 1999; i >= 673; --i) {
if (chk(i))
for (int j = 2; j < i && i + j < 2019; ++j) {
int k = 2019 - i - j;
if (i > j && j > k && chk(j) && chk(k)) ++cnt;
}
}
printf("%d\n", cnt); // 40785
return 0;
} 迷宫
01010101001011001001010110010110100100001000101010 00001000100000101010010000100000001001100110100101 01111011010010001000001101001011100011000000010000 01000000001010100011010000101000001010101011001011 00011111000000101000010010100010100000101100000000 11001000110101000010101100011010011010101011110111 00011011010101001001001010000001000101001110000000 10100000101000100110101010111110011000010000111010 00111000001010100001100010000001000101001100001001 11000110100001110010001001010101010101010001101000 00010000100100000101001010101110100010101010000101 11100100101001001000010000010101010100100100010100 00000010000000101011001111010001100000101010100011 10101010011100001000011000010110011110110100001000 10101010100001101010100101000010100000111011101001 10000000101100010000101100101101001011100000000100 10101001000000010100100001000100000100011110101001 00101001010101101001010100011010101101110000110101 11001010000100001100000010100101000001000111000010 00001000110000110101101000000100101001001000011101 10100101000101000000001110110010110101101010100001 00101000010000110101010000100010001001000100010101 10100001000110010001000010101001010101011111010010 00000100101000000110010100101001000001000000000010 11010000001001110111001001000011101001011011101000 00000110100010001000100000001000011101000000110011 10101000101000100010001111100010101001010000001000 10000010100101001010110000000100101010001011101000 00111100001000010000000110111000000001000000001011 10000001100111010111010001000110111010101101111000
#include <cstring>
#include <iostream>
#include <queue>
#define sc(x) scanf("%d", &(x))
const int ax[4] = {1, 0, 0, -1};
const int ay[4] = {0, -1, 1, 0};
const char dir[5] = {'D', 'L', 'R', 'U'};
int maze[105][105];
int X, Y;
using namespace std;
struct point {
int x, y;
string ans;
point(int a, int b, string c) : x(a), y(b), ans(c) {}
};
bool judge(int x, int y) {
if (x > 0 && x <= X && y > 0 && y <= Y && !maze[x][y]) return true;
return false;
}
void bfs() {
queue<point> ans;
ans.push(point(1, 1, ""));
while (!ans.empty()) {
point temp = ans.front();
ans.pop();
if (temp.x == X && temp.y == Y) {//抵达终点
cout << temp.ans << endl;
return;
}
for (int i = 0; i < 4; i++) {
int tempx = temp.x + ax[i];
int tempy = temp.y + ay[i];
if (judge(tempx, tempy)) {
maze[tempx][tempy] = 1;//不走回头路
ans.push(point(tempx,tempy,temp.ans+dir[i]));
}
}
}
}
int main() {
freopen("maze.txt","r",stdin);
X = 30, Y = 50;
for (int i = 1; i <= X; i++) {
for (int j = 1; j <= Y; j++) {
char t= getchar();
maze[i][j] = t-'0';
}
getchar();
}
bfs();
return 0;
} DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR
特别数的和
打表可以进行常数级别的优化。大约可以把时间降低到1/3到1/4。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
bool chk(int n) {
for (; n; n /= 10)
if (n % 10 == 2 ||n % 10 ==0||n % 10 ==1|| n % 10 == 9) return 1;
return 0;
}
int main() {
ll n;
while(~scanf("%lld",&n)) {
ll ans=0;
for(int i=1; i<=n; ++i)
if(chk(i))ans+=i;
printf("%lld\n",ans);
}
return 0;
} 完全二叉树的权值
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
const int N=1e5+7;
ll a[N];
int main() {
ll n;
while(~scanf("%lld",&n)) {
for(int i=1; i<=n; ++i)scanf("%lld",&a[i]);
ll ans=1,maxx=a[1],p=2,lv=1;
while(p<=n) {
++lv;
ll now=0;
for(int i=0; i<(1<<(lv-1))&&p<=n; ++i)now+=a[p++];
if(now>maxx)ans=lv,maxx=now;
}
printf("%lld\n",ans);
}
return 0;
} 等差数列
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
const int N=1e5+7;
ll a[N],d[N];
int main() {
ll n,m;
while(~scanf("%lld",&n)) {
for(int i=1; i<=n; ++i)scanf("%lld",&a[i]);
sort(a+1,a+1+n);
for(int i=1; i<n; ++i) d[i]=a[i+1]-a[i];
if(!d[1]) {
printf("%lld\n",n);
continue;
}
ll c=__gcd(d[1],d[2]);
for(int i=3; i<n; ++i) c=__gcd(c,d[i]);
ll ans=(a[n]-a[1])/c+1;
printf("%lld\n",ans);
}
return 0;
} 后缀表达式
#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
const int N = 2e5 + 7;
ll n, m, a[N];
int main() {
while (cin >> n >> m) {
for (int i = 0; i <= n + m; ++i) scanf("%lld", &a[i]);
sort(a, a + n + m + 1);
int L = 0, R = n + m;
ll ans = 0;
bool f = a[L] < 0 && m;
while (a[L] < 0 && m) {
if (m > 1 || a[L + 1] >= 0 && m == 1 || !n) --m;
else --n;
ans -= a[L++];
}
ans += a[R--];
while (R >= L && n) ans += a[R--], --n;
if (m) {
if (!f) ans -= a[L++], --m;
while (R >= L && m) ans += a[R--], --m;
}
printf("%lld\n", ans);
}
return 0;
} #include <algorithm>
#include <cstdio>
#include <iostream>
#define sc(x) scanf("%lld", &(x))
using namespace std;
typedef long long ll;
const int N = 2e5 + 7;
ll n, m, a[N];
int main() {
while (cin >> n >> m) {
ll sum = n + m + 1;
for (int i = 0; i < sum; ++i) sc(a[i]);
sort(a, a + sum);
int L = 0, R = sum - 1;
ll ans = 0;
bool f = 0;
while (a[L] < 0 && m) {
f = 1;
if (m > 1) // 当减号数量大于1的时候，优先用减号，将负数转化为正数
ans -= a[L++], --m;
else if (a[L + 1] >= 0 && m == 1 || !n) //如果下一个数是正数而且负号只剩一个
//那就用这个减号来把手上的负数变号
//或者正号耗尽 别无选择
ans -= a[L++], --m;
else if (n) //有减号的话，就可以把加号当减号用，将负数转化为正数
ans -= a[L++], --n;
}
ans += a[R--];
while (R >= L && n)
ans += a[R--], --n; //如果还有加号剩余，那么就用加号加大的数
if (m) { //如果还有减号
if (f) //如果之前对负数用过减号，那么就可以把减号当加号用，对正数操作
while (R >= L && m) ans += a[R--], --m;
else { //如果没有，就减去最小的，这样就可以负负得正了
ans -= a[L++], --m;
while (R >= L && m) ans += a[R--], --m;
}
}
printf("%lld\n", ans);
}
return 0;
} 除了以下情况，其他情况均为绝对值相加：
- 没有减号（全部相加）
- 全是负数（减去最大的值 其余相加）
- 全是正数（减去最小的值 其余相加）
灵能传输
混分代码
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
const int N=3e5+7;
ll a[N];
void dr(int i) {
a[i-1]+=a[i],a[i+1]+=a[i],a[i]=-a[i];
}
int main() {
ll n,T;
scanf("%lld",&T);
while(T--) {
scanf("%lld",&n);for(int i=1; i<=n; ++i)scanf("%lld",&a[i]);
for(int i=2;i<n;++i){
ll ori=max(abs(a[i-1]),abs(a[i+1]));
ll aft=max(abs(a[i-1]+a[i]),abs(a[i+1]+a[i]));
if(ori>aft)dr(i);
}
printf("%lld\n",*max_element(a+1,a+n+1));
}
return 0;
} 
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