T1

Problem

洛谷

Solution

这是线性扫描题吧。
就从1 ~ n 循环,若比起面高,则 ans += h[i] - h[i - 1]。

Code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
ll read()
{
    ll x = 0; int zf = 1; char ch;
    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') zf = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x * zf;
}
int x[100005];

int main()
{
    int n = read();
    x[0] = 0;
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        x[i] = read();
        if (x[i] > x[i - 1]) ans += x[i] - x[i - 1];
    }
    printf("%d\n", ans);
}

T2

Problem

洛谷

Solution

其实就是求一个数列里拐点数+1,然后O(n)扫一遍就好了。

Code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
ll read()
{
    ll x = 0; int zf = 1; char ch;
    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') zf = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x * zf;
}
int x[100005];

int main()
{
    int n = read();
    x[1] = read();
    int ansx = 1, ansy = 1;
    for (int i = 2; i <= n; i++)
    {
        x[i] = read();
        if (x[i] == x[i - 1]) continue;
        if (x[i] > x[i - 1]) ansx = max(ansx, ansy + 1);
        else ansy = max(ansy, ansx + 1);
    }
    printf("%d\n", max(ansx, ansy));
}
posted on 2017-11-04 23:09  WizardCowboy  阅读(91)  评论(0编辑  收藏  举报