牛客假日团队赛1 D.Promotion Counting

链接：

https://ac.nowcoder.com/acm/contest/918/D

题意：

Bessie the cow is helping Farmer John run the USA Cow Olympiad (USACO), an on-line contest where participants answer challenging questions to demonstrate their mastery of bovine trivia.

In response to a wider range of participant backgrounds, Farmer John recently expanded the contest to include four divisions of difficulty: bronze, silver, gold, and platinum. All new participants start in the bronze division, and any time they score perfectly on a contest they are promoted to the next-higher division. It is even possible for a participant to be promoted several times within the same contest. Farmer John keeps track of a list of all contest participants and their current divisions, so that he can start everyone out at the right level any time he holds a contest.

When publishing the results from his most recent contest, Farmer John wants to include information on the number of participants who were promoted from bronze to silver, from silver to gold, and from gold to platinum. However, he neglected to count promotions as they occurred during the contest. Bessie, being the clever bovine she is, realizes however that Farmer John can deduce the number of promotions that occurred solely from the number of participants at each level before and after the contest. Please help her perform this computation!

思路：

从最后往前叠加即可。

代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
const int MAXN = 3e5 + 10;
const int MOD = 1e9 + 7;
int n, m, k, t;

int main()
{
    int a1, a2, b1, b2, c1, c2, d1, d2;
    scanf("%d%d%d%d%d%d%d%d", &a1, &a2, &b1, &b2, &c1, &c2, &d1, &d2);
    int x, y, z;x = y = z = 0;
    x = d2-d1;
    y = c2-c1+x;
    z = b2-b1+y;
    printf("%d\n%d\n%d\n", z, y, x);

    return 0;
}