“玲珑杯”ACM比赛 Round #19

A -- A simple math problem

Time Limit：2s Memory Limit：128MByte

Submissions：1599Solved：270

DESCRIPTION

You have a sequence ${\mathrm{anan,\; which\; satisfies:}}^{}$

Now you should find the value of $\lfloor 10an\rfloor \lfloor 10an\rfloor .$

INPUT

The input includes multiple test cases. The number of test case is less than 1000. Each test case contains only one integer $\mathrm{n(1\le n\le 109)n(1\le n\le 109)。}$

OUTPUT

For each test case, print a line of one number which means the answer.

SAMPLE INPUT

5

20

1314

SAMPLE OUTPUT

5

21

1317

这个题就是找规律啊，一个很明显的规律是每一项都有一个贡献，但是这个范需要自己找，10的话按照log是要加1结果不需要，入手点就是这里，看看哪些少加了1，每个数量级多一个，这个pow损精度，wa了好几次，主要是n==1时输出值也是1，这个才是我被坑的最严重的

以下是题解

#include <stdio.h>
#include <math.h>
int pow10(int i){
int ans=1;
for(int j=0;j<i;j++)
    ans*=10;
return ans;
}
int main(){
int n;
while(~scanf("%d",&n)){
int ans=n+(int)log10(n*1.0);
if(n==10)ans-=1;
for(int i=2;i<=9;i++){
    int x=pow10(i);
    if(x+1-i<n&&n<x)
        ans++;
}
printf("%d
",ans);
}
return 0;}

B - Buildings

Time Limit：2s Memory Limit：128MByte

Submissions：699Solved：186

DESCRIPTION

There are $\mathrm{nn\; buildings\; lined\; up,\; and\; the\; height\; of\; the\; ii-th\; house\; is\; hihi.}$

An inteval $[l,r][l,r](l\le r)(l\le r)\; is\; harmonious\; if\; and\; only\; if\; max(hl,\dots ,hr)-min(hl,\dots ,hr)\le kmax(hl,\dots ,hr)-min(hl,\dots ,hr)\le k.$

Now you need to calculate the number of harmonious intevals.

INPUT

The first line contains two integers $\mathrm{n(1\le n\le 2\times 105),k(0\le k\le 109)n(1\le n\le 2\times 105),k(0\le k\le 109).\; The\; second\; line\; contains\; nn\; integers\; hi(1\le hi\le 109)hi(1\le hi\le 109).}$

OUTPUT

Print a line of one number which means the answer.

SAMPLE INPUT

3 1 1 2 3

SAMPLE OUTPUT

5

HINT

Harmonious intervals are: $[1,1],[2,2],[3,3],[1,2],[2,3][1,1],[2,2],[3,3],[1,2],[2,3].$

$模板题？？？区间最大值最小值的差小于等于k的值有多少组，单调栈，RMQ都不会被卡的$

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int maxN = 2e5+10;
int n, k, a[maxN];
LL ans;
void input() {
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; ++i)
        scanf("%d", &a[i]);
}
void work() {
    ans = 0;
    deque<int> mi, ma;
    int p = 0;
    for (int i = 0; i < n; ++i) {
        while (!(mi.empty() && ma.empty()) &&
                !(abs(a[i]-a[mi.front()]) <= k && abs(a[i]-a[ma.front()]) <= k)) {
            p++;
            while (!mi.empty() && mi.front() < p)
                mi.pop_front();
            while (!ma.empty() && ma.front() < p)
                ma.pop_front();
        }
        ans += i-p+1;
        while (!mi.empty() && a[mi.back()] > a[i])
            mi.pop_back();
        mi.push_back(i);
        while (!ma.empty() && a[ma.back()] < a[i])
            ma.pop_back();
        ma.push_back(i);
    }
    printf("%lld
", ans);
}

int main() {
    input();
    work();
    return 0;
}

 ST表

void rmqinit()  
{  
    for(int i = 1; i <= n; i++) mi[i][0] = mx[i][0] = w[i];  
    int m = (int)(log(n * 1.0) / log(2.0));  
    for(int i = 1; i <= m; i++)  
        for(int j = 1; j <= n; j++)  
        {  
            mx[j][i] = mx[j][i - 1];  
            if(j + (1 << (i - 1)) <= n) mx[j][i] = max(mx[j][i], mx[j + (1 << (i - 1))][i - 1]);  
            mi[j][i] = mi[j][i - 1];  
            if(j + (1 << (i - 1)) <= n) mi[j][i] = min(mi[j][i], mi[j + (1 << (i - 1))][i - 1]);  
        }  
}  
int rmqmin(int l,int r)  
{  
    int m = (int)(log((r - l + 1) * 1.0) / log(2.0));  
    return min(mi[l][m] , mi[r - (1 << m) + 1][m]);  
}  
int rmqmax(int l,int r)  
{  
    int m = (int)(log((r - l + 1) * 1.0) / log(2.0));  
    return max(mx[l][m] , mx[r - (1 << m) + 1][m]);  
}