"Blue Bridge Cup" 2017 Provincial Tournament Buns Counting-Mathematical Sieve Method DP

『Title description』

问题描述

　　小明几乎每天早晨都会在一家包子铺吃早餐。他发现这家包子铺有N种蒸笼，其中第i种蒸笼恰好能放Ai个包子。每种蒸笼都有非常多笼，可以认为是无限笼。
　　每当有顾客想买X个包子，卖包子的大叔就会迅速选出若干笼包子来，使得这若干笼中恰好一共有X个包子。比如一共有3种蒸笼，分别能放3、4和5个包子。当顾客想买11个包子时，大叔就会选2笼3个的再加1笼5个的（也可能选出1笼3个的再加2笼4个的）。
　　当然有时包子大叔无论如何也凑不出顾客想买的数量。比如一共有3种蒸笼，分别能放4、5和6个包子。而顾客想买7个包子时，大叔就凑不出来了。
　　小明想知道一共有多少种数目是包子大叔凑不出来的。

输入格式

　　第一行包含一个整数N。(1 <= N <= 100)
　　以下N行每行包含一个整数Ai。(1 <= Ai <= 100)

输出格式

　　一个整数代表答案。如果凑不出的数目有无限多个，输出INF。

样例输入

2
4
5

样例输出

6

样例输入

2
4
6

样例输出

INF

样例说明

　　对于样例1，凑不出的数目包括：1, 2, 3, 6, 7, 11。
　　对于样例2，所有奇数都凑不出来，所以有无限多个。

"Solution"

1. First solve the infinite case, if it is infinite, then only a part of the multiple is covered, and some are not covered, so if $g= gcd\{a[i]\},i=1,2,3...$ Is not $1$ , then these numbers only cover all $g$ multiples of$g$ , there must be multiples of a certain number that cannot be covered.
2. Use class idea similar to the linear sieve can be combined into the number of all screened out , a screen to set the upper limit of the number of first, a conclusion to use: for two prime numbers p, q, px + py not represent The maximum number is p * qpq. ( Proof link ) The maximum value of p and q is 100, then the upper limit can be determined as 1e4
3. Consider the setting of the sieve method: if a number $x$ can be combined, then $x-a[i]$ must also be able to be combined. With the idea of ​​a dp, all the numbers can be sieved out. The initial situation should be 0, (it can be considered that 0 $a[i]$ ). Many people write with complete backpacks, but the central idea of ​​the complete backpack in this question is justified.

/*****************************
*author:ccf
*source:POJ-
*topic:
*******************************/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#define ll long long
using namespace std;

const int N = 107,LIM = 1e4;
int n,m,cas;
int a[N];
bool bk[N*N];
ll gcd(ll a,ll b){
	return b ? gcd(b,a%b) : a;
}
int main(){
	//freopen("data.in","r",stdin);
	scanf("%d",&n);
	for(int i = 1; i <= n; ++i) scanf("%d",a+i);
	int g = a[1];
	for(int i  =2; i <= n; ++i) g = gcd(a[i],g);
	if(g != 1){
		printf("INF\n");
		return 0;
	}
	bk[0] = 1;
	for(int i = 1; i <= n; ++i){
		for(int j = a[i]; j < LIM; j++){
			bk[j] = max(bk[j],bk[j-a[i]]);
		}
	}
	int ans = 0;
	for(int i = 1; i < LIM; ++i) 
		if(!bk[i]) ans++;
	printf("%d\n",ans);
	return 0;
}