[2018 ICPC 青岛] 解题记录ing
M. Function and Function
队友说直接暴力即可
#include <cmath>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
int get(int x) {
if (x == 0) return 1;
int sum = 0;
while (x) {
int d = x % 10;
x /= 10;
if (d == 0 || d == 4 || d == 6 || d == 9) sum += 1;
else if (d == 8) sum += 2;
}
return sum;
}
int main() {
int t; scanf("%d", &t);
while (t--) {
int x, k; scanf("%d%d", &x, &k);
int res;
if (k == 0) {
res = x;
printf("%d\n", res);
continue;
}
while (k--) {
x = get(x);
res = x;
if (x == 1) {
if (k % 2 == 0) {
res = 1;
}
else res = 0;
break;
}
}
printf("%d\n", res);
}
return 0;
}
C.
const int maxn=1e6+7;
int T,n,a[maxn],b[maxn],f;
char x[maxn],y[maxn];
ll sum;
int main(){
T=read();
while(T--){
n=read();
scanf("%s%s",x,y);
f=0;
for(int i=0;i<n;i++){
a[i+1]=x[i]-'0';
b[i+1]=y[i]-'0';
if(a[i]==b[i]&&a[i+1]!=b[i+1]) f++;
}
if(n==1){
if(a[1]!=b[1]) sum=0;
else sum=1;
out(sum); putchar('\n');
continue;
}
if(f==0) sum=1ll*n*(n+1)/2;
else if(f==1) sum=(n-1)*2;
else if(f==2) sum=6;
else sum=0;
out(sum); putchar('\n');
}
}
Plants vs. Zombies
二分答案
在处理的过程中直接在下一个和当前位置左右横跳即可
#define Clear(x,val) memset(x,val,sizeof x)
ll a[maxn];
ll nowCnt[maxn],needCnt[maxn];
ll n,m;
bool check(ll mid) {
if(mid == 0LL) return true;
for(int i=1; i<=n; i++) {
nowCnt[i] = 0;
needCnt[i] = (mid - 1) / a[i] + 1;
}
for(int i=1; i<=n; i++) {
if(n == i && nowCnt[i] >= needCnt[i]) return true;
/**
if(nowCnt[i] >= needCnt[i]) {
if(i == n) return true;
// continue;
}
**/
if(m <= 0) return false;
++ nowCnt[i];
-- m;
if(nowCnt[i] >= needCnt[i]) continue;/// already okay
///else:
ll stillNeed = needCnt[i] - nowCnt[i];
if(stillNeed*2 > m) return false;
m -= 2 * stillNeed;
nowCnt[i+1] += stillNeed;/// * 2
/**
cur_pos + 1
**/
}
return true;
}
int main() {
int _ = read;
while (_ --) {
n = read,m = read;
ll tm = m;
ll mini = 0x3f3f3f3f;
for(int i=1; i<=n; i++) {
a[i] = read;
mini = min(mini,a[i]);
}
// /**
if(m == 0) {
puts("0");
} else if(m < n) {
puts("0");
} else if(m == n) {
printf("%lld\n",mini);
} else {
// **/
ll l = 0,r = 1e17 + 1;
while(l < r) {
ll mid = l + r + 1>> 1;
m = tm;
if(check(mid)) l = mid;
else r = mid - 1;
}
printf("%lld\n",l);
}
}
return 0;
}
/**
2
4 8
3 2 6 6
3 9
10 10 1
**/
J . Books
二分了好久发现不是二分,比较考验思维
附带几组hack 数据样例
int a[maxn];
vector<int> vet;
int main() {
int _ = read;
while (_ --) {
int n = read,m = read;
int cnt = 0;
vet.clear();
vet.push_back(0);
int zero = 0;
for (int i = 1; i <= n; i++) {
scanf("%lld", &a[i]);
if (a[i] == 0) zero++;
else vet.push_back(a[i]);
}
if (zero > m) {
puts("Impossible");
continue;
}
if (n == m) {
puts("Richman");
continue;
}
int mini = 0x3f3f3f3f;
int siz = vet.size() - 1;
m -= zero;
ll ans = 0;
int select = 0;
for(int i=1; i<=siz; i++) {
++ select;
if(select <= m) {
ans += vet[i];
} else {
mini = min(mini,vet[i]);
}
}
// cout << ans << endl;
// cout << mini << endl;
ans += mini - 1;
// cout << ans << endl;
printf("%lld\n",ans);
}
return 0;
}
/**
2
1
5 4
0 0 0 0 1
1
10 3
100000 100000 100000 1 1 1 1 1 1 1
1
10 6
10000 1 1 1 1 1 1 1 10000 10000
4
4 2
1 2 4 8
4 0
100 99 98 97
2 2
10000 10000
5 3
0 0 0 0 1
**/
I. Soldier Game
三维线段树
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod = (int)1e9+7;
const int N = 1e5 + 100;
int a[N][2];
pll p[N<<1];
bool cmp(pll & x, pll & y){
return a[x.fi][x.se] < a[y.fi][y.se];
}
int tree[N<<2][2][2];
void Build(int l, int r, int rt){
tree[rt][0][0] = tree[rt][0][1] = tree[rt][1][0] = tree[rt][1][1] = 0;
if(l == r) {
tree[rt][1][0] = 1;
return ;
}
int m = l+r >> 1;
Build(lson); Build(rson);
}
void PushUp(int rt){
tree[rt][0][0] = (tree[rt<<1][0][0] && tree[rt<<1|1][0][0]) || (tree[rt<<1][0][1] && tree[rt<<1|1][1][0]);
tree[rt][0][1] = (tree[rt<<1][0][0] && tree[rt<<1|1][0][1]) || (tree[rt<<1][0][1] && tree[rt<<1|1][1][1]);
tree[rt][1][0] = (tree[rt<<1][1][0] && tree[rt<<1|1][0][0]) || (tree[rt<<1][1][1] && tree[rt<<1|1][1][0]);
tree[rt][1][1] = (tree[rt<<1][1][1] && tree[rt<<1|1][1][1]) || (tree[rt<<1][1][0] && tree[rt<<1|1][0][1]);
}
void Update(int L, int op,int l, int r, int rt){
if(l == r){
tree[rt][0][op] ^= 1;
return ;
}
int m = l+r >> 1;
if(L <= m) Update(L, op, lson);
else Update(L, op, rson);
PushUp(rt);
}
int main(){
int T, n;
scanf("%d", &T);
while(T--){
scanf("%d", &n);
int m = 0;
for(int i = 1; i <= n; ++i){
scanf("%d", &a[i][0]);
p[++m] = pll(i,0);
if(i-1) {
a[i-1][1] = a[i-1][0] + a[i][0];
p[++m] = pll(i-1,1);
}
}
sort(p+1, p+1+m, cmp);
Build(1,n,1);
LL ans = INF;
for(int i = 1, j=1; i <= m; ++i){
while(j <= m && !tree[1][0][0]){
Update(p[j].fi, p[j].se, 1, n, 1);
j++;
}
if(tree[1][0][0]) ans = min(ans, 1ll*a[p[j-1].fi][p[j-1].se] - a[p[i].fi][p[i].se]);
Update(p[i].fi, p[i].se, 1, n, 1);
}
printf("%lld\n", ans);
}
return 0;
}