CodeForces - 617B (思维题）

Bob loves everything sweet. His favorite chocolate bar consists of pieces, each piece may contain a nut. Bob wants to break the bar of chocolate into multiple pieces so that each part would contain exactly one nut and any break line goes between two adjacent pieces.

You are asked to calculate the number of ways he can do it. Two ways to break chocolate are considered distinct if one of them contains a break between some two adjacent pieces and the other one doesn't.

Please note, that if Bob doesn't make any breaks, all the bar will form one piece and it still has to have exactly one nut.

题意：就是一串01串（可能不含0或1），把该串分成多份，使每份含有一个1，为有几种分法

只要求出每两个1的下标差，再全部乘起来就好了

#include <vector>
#include <stdio.h>
#include <map>
#include <stdlib.h>
#include<ctype.h>
#include <iostream>
#include <queue>
#define LL long long
using namespace std;

const int MAX = 1e6 + 50;

int a[105];
vector<int> vec; // 存每个1的下标
int main(int argc, char const *argv[]){	
	int n;
	scanf("%d", &n);
	int cnt = 0;
	for(int i = 0; i < n; i++){
		int x;
		scanf("%d", &x);
		if(x){
			cnt++;
			vec.push_back(i);
		}
	}

	int len = vec.size();

	if(cnt == 1){
		printf("1\n");
	} else if(cnt == 0){
		printf("0\n");
	} else{
		LL ans = 1;
		for(int i = 1; i < len; i++){
			ans *= ((LL)vec[i] - (LL)vec[i - 1]);
		}
		printf("%lld\n", ans);
	}
	return 0;
}