AtCoder Beginner Contest 259 - A - Growth Record

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $100$ points

Problem Statement

Takahashi had his $N$-th birthday, when he was $T$ centimeters tall.
Additionally, we know the following facts:

• In each year between Takahashi's birth ($0$-th birthday) and his $X$-th birthday, his height increased by $D$ centimeters. More formally, for each $i = 1, 2, \ldots, X$, his height increased by $D$ centimeters between his $(i-1)$-th birthday and his $i$-th birthday.
• Between Takahashi's $X$-th birthday and his $N$-th birthday, his height did not change.

Find Takahashi's height on his $M$-th birthday, in centimeters.

Constraints

• $0 \leq M \lt N \leq 100$
• $1 \leq X \leq N$
• $1 \leq T \leq 200$
• $1 \leq D \leq 100$
• Takahashi was at least $1$ centimeter tall at his birth.
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$ $M$ $X$ $T$ $D$

Output

Print the answer as an integer.

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Sample Input 1

38 20 17 168 3

Sample Output 1

168

In this sample, Takahashi was $168$ centimeters tall on his $38$-th birthday. Also, his height did not change between his $17$-th birthday and $38$-th birthday.
From these facts, we find that he was $168$ centimeters tall on his $20$-th birthday, so the answer is $168$.

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Sample Input 2

1 0 1 3 2

Sample Output 2

1

In this sample, Takahashi was $1$ centimeter tall on his $0(=M)$-th birthday and $3(=T)$ centimeters tall on his $1(=N)$-st birthday.

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Sample Input 3

100 10 100 180 1

Sample Output 3

90

题目大意

高橋君$N$岁时身高为$T$，在$X$岁之前，每年长$D$厘米。（$X\sim N$岁就不长了）

问高橋君$M$岁时多高。

解题思路

题目保证了$X < N$，也就是说现在高橋君的身高就是他的最终身高（已经不长了）

因此，对于询问的$M$，如果$M\geq X$，那么就输出高橋君的最终身高；否则看从$M$到$X$有几年，就说明长了几个$D$，用最终身高减去$X-M$个$D$即可。

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AC代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;
int main() {
    int n, m, x, t, d;
    cin >> n >> m >> x >> t >> d;
    if (m >= x) {
        cout << t << endl;
    }
    else {
        int diffYear = x - m;
        cout << t - d * diffYear << endl;
    }
    return 0;
}

同步发文于CSDN，原创不易，转载请附上原文链接哦~
Tisfy：https://letmefly.blog.csdn.net/article/details/125699180