Codeforces Educational Codeforces Round 3 B. The Best Gift（签到）-优快云博客

本文探讨了在给定数量的书籍和限定类型的条件下，如何利用组合数学原理计算不同类型的书籍两两配对的方法总数。通过具体公式与代码实现，展示了组合数学的实际应用。

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题意:

$给定n\le 10^5本书, 类型数\le 10, 每次选两种不同的类型, 多少种选法$

分析:

$ans = \sum all - \sum same\ genre = C_n^2 - \sum_{m=1}^{10}C_{a_m}^2$
$结果不爆int, 不代表过程中不爆$

代码:

//
//  Created by TaoSama on 2015-12-19
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

typedef long long LL;
LL n, m, cnt[15];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d%d", &n, &m) == 2) {
        memset(cnt, 0, sizeof cnt);
        for(int i = 1; i <= n; ++i) {
            int x; scanf("%d", &x);
            ++cnt[x];
        }
        LL ans = n * (n - 1) >> 1;
        for(int i = 1; i <= m; ++i)
            ans -= cnt[i] * (cnt[i] - 1) >> 1;
        printf("%I64d\n", ans);
    }
    return 0;
}