“玲珑杯”ACM比赛 Round #19

1001

打表猜了一下规律....
#include <cstdio>
#include <cstring>
#include <cmath>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mst(Arr, x) memset(Arr, x, sizeof Arr)
#define REP(i, x, n) for(int i = x; i < n; ++i)
const int qq = 1e6 + 10;
const int MOD = 1e9 + 7;
int n;
double ans[qq];
int num[] = {11, 100, 999, 9998, 99997, 999996, 9999995, 99999994, 999999993};

int main(){
/* ans[1] = 0;
for(int i = 2; i < qq; ++i) {
ans[i] = (log10(i * 1.0 + ans[i - 1]));
}*/
while(scanf("%d", &n) != EOF) {
/* for(int i = 1; i <= n; ++i) {
printf("%d\n", (int)floor(pow(10.0, ans[i])));
}*/
int t = 0;
for(int i = 0; i < 9; ++i) {
if(n >= num[i]) t++;
}
printf("%d\n", n + t);
}
return 0;
}
1002

题意：求有多少区间 ，区间的最大值 - 区间的最小值 <= k

思路：如果区间[l, r]，满足那么在[l, r]区间组成的小区间也一定满足条件，所以我们确定r的位置，二分找最远的l，那么以r为区间右端满足的就有r - l + 1，区间最值我们用RMQ处理就好

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <utility>
#include <queue>
#include <stack>
#include <string>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define ft first
#define sd second
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int qq = 2e5 + 10;
const int MOD = 1e9 + 7;
int maxn[qq][20], minx[qq][20];
int num[qq];
int n, k;
void RmqInit() {
for(int i = 1; i <= n; ++i) {
maxn[i][0] = minx[i][0] = num[i];
}
for(int j = 1; (1 << j) <= n; ++j) {
for(int i = 1; i + (1 << j) - 1 <= n; ++i) {
maxn[i][j] = max(maxn[i][j - 1], maxn[i + (1 << (j - 1))][j - 1]);
minx[i][j] = min(minx[i][j - 1], minx[i + (1 << (j - 1))][j - 1]);
}
}
}
bool Check(int l, int r) {
int t = 0;
while((1 << (t + 1)) <= r - l + 1) t++;
int a = max(maxn[l][t], maxn[r - (1 << t) + 1][t]);
int b = min(minx[l][t], minx[r - (1 << t) + 1][t]);
if(a - b <= k) return true;
return false;
}

LL Solve(int x) {
int l = 1, r = x;
LL ans = -1;
while(l <= r) {
int mid = (l + r) / 2;
if(Check(mid, x)) {
ans = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
if(ans == -1) return 0;
return x - ans + 1;
}

int main(){
while(scanf("%d%d", &n, &k) != EOF) {
for(int i = 1; i <= n; ++i) {
scanf("%d", num + i);
}
RmqInit();
LL ans = 0;
for(int i = 1; i <= n; ++i) {
ans += Solve(i);
}
printf("%lld\n", ans);
}
return 0;
}