P2327

code[class*="language-"] {
padding: .1em;
border-radius: .3em;
white-space: normal;
background: #f5f5f5;
}

.token.comment,
.token.blockquote {
color: #969896;
}

.token.cdata {
color: #183691;
}

.token.doctype,
.token.punctuation,
.token.variable,
.token.macro.property {
color: #333;
}

.token.operator,
.token.important,
.token.keyword,
.token.rule,
.token.builtin {
color: #a71d5d;
}

.token.string,
.token.url,
.token.regex,
.token.attr-value {
color: #183691;
}

.token.property,
.token.number,
.token.boolean,
.token.entity,
.token.atrule,
.token.constant,
.token.symbol,
.token.command,
.token.code {
color: #0086b3;
}

.token.tag,
.token.selector,
.token.prolog {
color: #63a35c;
}

.token.function,
.token.namespace,
.token.pseudo-element,
.token.class,
.token.class-name,
.token.pseudo-class,
.token.id,
.token.url-reference .token.variable,
.token.attr-name {
color: #795da3;
}

.token.entity {
cursor: help;
}

.token.title,
.token.title .token.punctuation {
font-weight: bold;
color: #1d3e81;
}

.token.list {
color: #ed6a43;
}

.token.inserted {
background-color: #eaffea;
color: #55a532;
}

.token.deleted {
background-color: #ffecec;
color: #bd2c00;
}

.token.bold {
font-weight: bold;
}

.token.italic {
font-style: italic;
}

/* JSON */
.language-json .token.property {
color: #183691;
}

.language-markup .token.tag .token.punctuation {
color: #333;
}

/* CSS */
code.language-css,
.language-css .token.function {
color: #0086b3;
}

/* YAML */
.language-yaml .token.atrule {
color: #63a35c;
}

code.language-yaml {
color: #183691;
}

/* Ruby */
.language-ruby .token.function {
color: #333;
}

/* Markdown */
.language-markdown .token.url {
color: #795da3;
}

/* Makefile */
.language-makefile .token.symbol {
color: #795da3;
}

.language-makefile .token.variable {
color: #183691;
}

.language-makefile .token.builtin {
color: #0086b3;
}

/* Bash */
.language-bash .token.keyword {
color: #0086b3;
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.code-chunk .btn-group{display:none}.markdown-preview:not([for="preview"]) .code-chunk .status{display:none}.markdown-preview:not([for="preview"]) .code-chunk .output-div{margin-bottom:16px}.scrollbar-style::-webkit-scrollbar{width:8px}.scrollbar-style::-webkit-scrollbar-track{border-radius:10px;background-color:transparent}.scrollbar-style::-webkit-scrollbar-thumb{border-radius:5px;background-color:rgba(150,150,150,0.66);border:4px solid rgba(150,150,150,0.66);background-clip:content-box}html body[for="html-export"]:not([data-presentation-mode]){position:relative;width:100%;height:100%;top:0;left:0;margin:0;padding:0;overflow:auto}html body[for="html-export"]:not([data-presentation-mode]) .markdown-preview{position:relative;top:0}@media screen and (min-width:914px){html body[for="html-export"]:not([data-presentation-mode]) .markdown-preview{padding:2em calc(50% - 457px)}}@media screen and (max-width:914px){html body[for="html-export"]:not([data-presentation-mode]) 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html .markdown-preview.markdown-preview {
font-family: 'PingFang SC', 'Consolas', 'Microsoft YaHei', sans-serif, monospace;
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font-family: 'PingFang SC';
src: url('file:///c:/windows/fonts/PingFang Regular.ttf');
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font-family: 'Consolas', 'PingFang SC', 'Microsoft YaHei', sans-serif, monospace;
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html .markdown-preview.markdown-preview h1,
html .markdown-preview.markdown-preview h2,
html .markdown-preview.markdown-preview h3,
html .markdown-preview.markdown-preview h4,
html .markdown-preview.markdown-preview h5,
html .markdown-preview.markdown-preview h6 {
font-family: 'PingFang SC', 'Consolas', 'Microsoft YaHei', monospace, sans-serif;
}

-->

P2327 [SCOI2005]扫雷

题目描述

输入输出格式

输入格式：

第一行为N，第二行有N个数，依次为第二列的格子中的数。（1<=N<=10000）

输出格式：

一个数，即第一列中雷的摆放方案数。

输入输出样例

输入样例#1：

2
1 1

输出样例#1：

做法:

题解区大家都是八仙过海.
我想了一种比较恶心但是能过的方法.比题解区某些做法要好一点点.

首先需要分析一下这个题目,经过一番脑洞之后发现答案必定大于0小于等于2 根据这个原理我们就可以大力模拟或者是搜索.
当然也可以 dp .万物皆可dp

然后这个做法就是 dp 做法.

$$f[i][j][k]表示当前位置为 i ,当前位置是不是雷(用j表示),下一位置需不需要是雷(k)的方案数.$$

转移就根据上一位置的选择.

#include<iostream>

#include<cstdio>

#define N 10005

using namespace std;

void read(int &s){

    char ch=getchar();

    for(;!isdigit(ch);ch=getchar());

    for(s=0;isdigit(ch);s=s*10+ch-'0',ch=getchar());

}

int n;

int g[N];

int f[N][2][2];

int main(){

    read(n);

    for(int i=1;i<=n;++i)read(g[i]);

    f[0][0][0]=f[0][0][1]=1;

    for(int i=1;i<=n;++i){

        if(g[i]==0){

            f[i][0][0]+=f[i-1][0][0];

        }

        if(g[i]==1){

            f[i][0][0]+=f[i-1][1][0];

            f[i][1][0]+=f[i-1][0][1];

            f[i][0][1]+=f[i-1][0][0];

        }

        if(g[i]==2){

            f[i][1][1]+=f[i-1][0][1];

            f[i][0][1]+=f[i-1][1][0];

            f[i][1][0]+=f[i-1][1][1];

        }

        if(g[i]==3)

            f[i][1][1]+=f[i-1][1][1];

    }

    printf("%d",f[n][0][0]+f[n][1][0]);

    return 0;

}

P2327 [SCOI2005]扫雷

≡

巴特西