兰州交通大学硕士论文字数要求不低于300字,不高于500字。
首先要说明的一点是每个学样对论文都会有一定的要求 虽然大体上不会有太大差异但在字体、字号等方面还是不同的 所以一定要以你所要毕业的学校要求为准。
下面是摘录的别的学样的要求
学生毕业设计(论文)书写格式要求
(说明:以下所有红色、蓝色文字仅供参考,学生在写作论文时请保留字体、字号,改写或删除掉文字,黑色文字请保留。论文统一用A4纸打印,页面设置:上边距25mm,左边距25mm,右边距20mm,下边距20mm;页眉16mm,页脚16mm;全文行间距统一为1.25行。页眉内容统一为:“安徽科技学院 食品药品学院 本科毕业论文”,字体为仿宋,小五号,分两边居中。为保证打印效果,在打印前,请将全文字体的颜色统一设置成黑色。
一、论文封面(采用×××学校毕业论文统一格式);
二、目录:小四号字、宋体(建议在一页纸上完成);
三、标题:小二号字、黑体,段前段后行距为0.5行,居中;
四、学生和指导教师姓名:小四号字、宋体,段前段后行距为0.5行,居中;
五、中文摘要、关键词:五号字、楷体,左右各缩进两个字距,标题本身加粗;
六、一级标题:四号字、黑体,段前段后行距为0.5行;
七、二级标题:小四号字、黑体,段前段后行距为0.5行;
八、三级标题:小四号字、宋体加粗(四级标题以下按正文字体处理,加粗);
九、正文:小四号字、宋体;
十、参考文献:五号字、宋体;
十一、英文标题:最多不超过两行,三号字,Times New Roman字体;
十二、英文摘要、关键词:五号字、Times New Roman字体;
十三、论文综述、开题报告等所有资料都参考本格式版式。
(正文格式详见下页)
(顶头空2行)目 录(四号黑体,居中)
摘要……………………………………………………………………………………………1
关键词…………………………………………………………………………………………1
前言(或引言)………………………………………………………………………………1
1□材料与方法………………………………………………………………………………Y
1.1□材料 ……………………………………………………………………………………Y
1.2□方法 ……………………………………………………………………………………Y
1.2.1□×××××…………………………………………………………………………Y
1.2.2□×××××…………………………………………………………………………Y
1.2.3□×××××…………………………………………………………………………Y
1.2.4□×××××…………………………………………………………………………Y
2□××………………………………………………………………………………………Y
2.1□×××××……………………………………………………………………………Y
3□×××…………………………………………………………………………………… Y
……………………………………………………………(略)
致谢……………………………………………………………………………………………Y
参考文献………………………………………………………………………………………Y
注:1. 目次中的内容一般列出“章”、“节”、“条”三级标题即可;
2.X、Y表示具体的阿拉伯数字;
(题 目)
某某专业:×××(学生姓名)
指导老师:×××(指导教师姓名)
摘 要:××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××。
关键词:×××, ×××, ×××, ×××, ×××
前言(一级标题)
××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××(正文)
1 材料与方法(一级标题)
1.1 试剂与仪器(二级标题)
1.1.1 试剂(三级标题)
氢氧化钠、×××××、×××××、×××××××××××××××××××××××××××××××××××(正文)
1.1.2 仪器(三级标题)
分析天平、×××××、×××××、×××××、××××××××××××××××××××××××××××××【1】(正文)
×××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××
………
致谢:本次实验××××××××××××××××××××××××××××××××××××××××。
参 考 文 献
[1] 作者姓名,作者姓名.参考文献题目[J]. 期刊或杂志等名称,年份,卷(期数):页码.
[2] 刘凡丰. 美国研究型大学本科教育改革透视[J] . 高等教育研究,2003,5(1):18-19.
没有卷的就直接写2003(1)(本条为期刊杂志著录格式)
[3] 谭丙煜.怎样撰写科学论文[M].2版.沈阳:辽宁人民出版社,1982:5-6.(本条为中文图书著录格式)
[4] 作者姓名. 参考文献题目[D].南京:南京农业大学,2002:页码.(本条为硕士、博士论文著录格式)
[5] 作者姓名. 参考文献题目[N].人民日报,2005-06-12.(本条为报纸著录格式)
[6] 作者姓名. 参考文献题目[C]// 作者姓名.论文集名称.城市:出版单位(社),年代:页码.(本条为论文集著录格式)
[7] 外国作者姓名. 参考文献题目[M].译者(名字),译.城市:出版单位,年代:页码.(本条为原著翻译中文的著录格式,多个译者可写为:***,***,***,等译.)
外文文献著录格式参照中文的(五号Times New Romar)。
文献类型标志说明:普通图书 M ,会议记录C,汇编G,报纸N,期刊J,学位论文D,报告R,标准S,专利P,数据库DB,计算机程序CP。
×××××××××(Title, 三号Times New Romar)
××× (Student name),××× (Tutor name)(小四号Times New Romar)
School of Food and Drug, Anhui Science and Technology University, Fengyang 233100, Anhui, China
Abstract: ×××××××××(五号Times New Romar,150-300个实词)××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××
Key words: ×××; ×××; ×××; ×××(3-6实词个,五号Times New Romar)
论文中图的具体要求为:
① 主线粗于辅线(座标线)
② 图题,小5黑(句末无标点)
③ 标值线(座标上的刻度线)一律在图的内侧
④ 图例一律在图题的上方或在图中,6宋
⑤ 图注一律在图题的下方,6宋
⑥ 标目(座标的文字说明)及图内文字,6宋
⑦ 图版(照片)说明在图题之下,6宋,文字一般接排,如:A.麦穗形态;B.花原基
论文中表格的具体要求为:
(注:表格要用三线格,图表标题要翻译成英文)
① 表题:小5黑,居中(句末无标点)
② 表内容:6宋
③ 数字一般以小数点位数对齐,数值后表示差异显著性的字母右肩上标
④ 表注:6宋,各注之间用“;”隔开
楼上说的似乎都太小儿科了,楼主想必是要发表的那种,当然要正式一点.
这里的一篇是偏向交作业的
下面一个是正式发表的双语版本
张彧典人工证明四色猜想 山西盂县党校数学高级讲师
用25年业余时间研究四色猜想的人工证明。在借鉴肯普链法和郝伍德范例正反两方面做法的基础上,独创了郝——张染色程序和色链的数量组合、位置(相交)组合理论,确立了仅包含九大构形的不可免集合,从而弥补了肯普证明中的漏洞。现贴出全文(中——英文对照)及参考文献的英译汉全文。欢迎各位同仁批评指正。
最后特别感谢英国兰开斯特大学A.lehoyd、兰州交大张忠辅、清华大学林翠琴、上海师大吴望名四位教授的无私帮助。
附:论文
用“H·Z—CP“求解赫伍德构形
张彧典 (山西省盂县县委党校 045100)
摘要:本文根据色链的数量和位置组合理论,用赫伍德染色程序(简称H—CP)和张彧典染色程序(简称Z—CP)找到一个赫伍德构形的不可避免集。
关键词:H—CP Z—CP H·Z—CP
《已知的赫伍德范例》〔1〕对求解赫伍德构形有两大贡献。其一,提供了H—CP,使我们用它找到了赫伍德染色非周期转化的赫伍德构形组合;其二,范例2提供了赫伍德染色周期转化的赫伍德构形,使我们发现了Z—CP,解决了这种构形的正确染色。
为下面讨论方便,先给出〔1〕文中赫伍德构形的最简单模型。
如图1所示:
四色用A、B、C、D表示,待染色区V用小圆表示,其五个邻点染色用A1、B1、B2、C1、D1表示,形成的五边形区域叫双B夹A型中心区。中心区外有A1—C1链、A1—D1链(因它们的首尾分别被V连成环,故叫环,以便与开放链区分),其中还有B1—D2链、B2—C2链,A1、A2被C2—D2链隔开。其余赫伍德构形类同。
在我们所设的模型中,再添加一些不同的色链后就构成许多不同的标准三角剖分图(记为G′)。当借助H—CP对它们求解时发现,其中色链的不同数量组合和相交组合直接影响解法上的差异。
现在具体确立赫伍德构形的不可避免集。
在后面图解中,画小横线者表示环,画粗线者表示两点以上染色互换的链,B(D)等表示一个点的染色互换。
如图2: 设图1中有B1-A2链、D1-C2链(也可以是B2-A2链)存在时。
其解法是:在A1—C1环内作B、D互换,生成新的A—D环(生不成情形归于下一种构形),再作A—D环外的C、B互换,可给V染C色。
如图3:设图1中有C1-D2链、D1-C2链存在时。
其解法是:在A1—C1环内作B、D互换,生成B—C环;作B—C环外的D、A互换,生成新的A—C环(生不成情形归于下一种构形);再作A—C环内的B、D互换,可给V染B色。
如图4:设图1中有C1-D2链、B2-A2链存在时。
其解法是:在A1—C1环内作B、D互换,生成B—C环;作B—C环外的D、A互换,生成B—D环;作B—D环内的A、C互换,生成新的B—C环(生不成情形归于下一种构形);再作B—C环内的D、A互换,可给V染D色。
如图5:设图4中B1-D2链与A1-D1环相交,这时有B1-A3、C1-A3生成。
其解法是:在A1—C1环内作B、D互换,生成B—C环;作B—C环外的D、A互换,生成B—D环;作B—D环内的A、C互换,生成A—D环;作A—D环外的C、B互换,生成新的B—D环(生不成情形归于下一种构形);再作B—D环外的A、C互换,可给V染A色。
如图6:设图5中C1-D2链与A1-C1环相交,为简单起见,将C1-D2链在A1-C1环外的D色点均改染B色,见图中B(带圈子的)。
其解法是:在A1—C1环内作B、D互换,生成B—C环;作B—C环外的D、A互换,生成B—D环;作B—D环内的A、C互换,生成A—D环;作A—D环外的C、B互换,生成A—C环;作A—C环外的B、D互换,生成新的A—D环(生不成情形归于下一种构形);再作A—D环内的C、B互换,可给V染C色。
如图7:设图6中B1-D2链再与B1-A3链相交,为简单起见,将B1-A3链在B1-D2链内侧的A色点均改染C色,见图中C(带圈子的)。
其解法是:在A1—C1环内作B、D互换,生成B—C环;作B—C环外的D、A互换,生成B—D环;作B—D环内的A、C互换,生成A—D环;作A—D环外的C、B互换,生成A—C环;作A—C环外的B、D互换,生成B—C环;作B—C环内的D、A互换生成新的A—C环(生不成情形归于下一种构形);再作A—C环内的B、D互换,可给V染B色。
如图8:设图7中有B1-D2链与C1-D2链在A1-C1环内相交。
其解法是:在A1—C1环内作B、D互换,生成B—C环;作B—C环外的D、A互换,生成B—D环;作B—D环内的A、C互换,生成A—D环;作A—D环外的C、B互换,生成A—C环;作A—C环外的B、D互换,生成B—C环;作B—C环内的D、A互换生成B—D环;作B—D环外的A、C互换,生成新的B—C环(生不成情形归于下一种构形);再作B—C环内的D、A互换,可给V染D色。
图9:设图8中有B2-A2链与A1-D1环相交。
其解法是:在A1—C1环内作B、D互换,生成B—C环;作B—C环外的D、A互换,生成B—D环;作B—D环内的A、C互换,生成A—D环;作A—D环外的C、B互换,生成A—C环;作A—C环外的B、D互换,生成B—C环;作B—C环内的D、A互换生成B—D环;作B—D环外的A、C互换,生成A—D环;作A—D环内的C、B互换,生成新的B—D环;(生不成情形归于下一种构形)再作B—D环内的A、C互换,可给V染A色。
如图10:这是一个十折对称的赫伍德构形。即在图3中,按图6的相交组合方式设C1—D2链与A1—C1环相交,D1—C2链与A1—D1环相交,C1—D2链在A1—C1环外的D色点与D1—C2链在A1—D1环外的C色点均改染B色,见图中B(带圈子的)。;再设改染成的C—B链、D—B链对称相交。这个赫伍德构形就是〔1〕文中范例2的拓扑变换形式。
对于图10如果沿用图2—9的求解方法,就会产生四个周期转化的赫伍德构形,无法得解。但是,四个连续转化的赫伍德构形有一个共同的染色特征,即都包含A—B环,于是产生了如下特殊的Z—CP:
若已知的是第一(或三)图时,先作A—B环外的C,D互换,生成新的A—C,A—D(或B—C、B—D)环,再作B(D)、B(C)[或A(D)、A(C)]互换,使五边形五个顶点染色数减少到3。解如图10(1)和图10(3)。
若已知的是第二(或四)图时,先作A—B环外的C,D互换,生成了新的B—C(或A—D)链,再作B—C(或A—D)链一侧的A(D)[或A(C)〕互换,使五边形五个顶点染色数减少到3。解如图10(2)和10(4)。
下面从理论上证明图2—10组成的不可避免集的完备性。
在已四染色的G’中,由A、B、C、D四色中任意二色组成的不同色链共C42(=6) 种。反映在赫伍德构形中,有始点终点均在中心区且相交的A1-C1环、A1-D1环,还有始点在中心区,终点在A1-C1、A1-D1二环交集区域边缘上的B1-D2、B1-A2(B2-A2)、B2-C2、C1-D2(D1-C2)四种链。这四种链在赫伍德构形中的不同数量组合共四组:
B1-A2、B1-D2、B2-C2、B2-A2
B1-A2、B1-D2、B2-C2、D1-C2
C1-D2、B1-D2、B2-C2、B2-A2
C1-D2、B1-D2、B2-C2、D1-C2
而六种色链中任意两种色链的不同位置组合共C62(=15)组。其中有三组不可相交组合:
A-B与C-D、A-C与B-D、A-D与B-C;
还有12组可相交组合:
A-B与A-C、A-D、B-C、B-D;
A-C与A-D、B-C、C-D ;
A-D与B-D、C-D;
B-C与B-D、C-D;
B-D与C-D。
我们把上述六种色链的不同数量组合(4组)及不同位置组合(12组可相交的)作为两大变量,一共可得到16种不同组合的赫伍德构形;然后在“结构最简”和“解法相同”的约束条件下逐一检验,具体归纳为:图2——4体现四种不同数量组合,其中图2体现前两种组合;图5——9体现依次增多的相交组合,其中图9已包含了12种相交组合;图10体现特殊的数量组合和相交组合。
到此,我们用“H·Z—CP”成功地解决了赫伍德构形的正确染色,从而弥补了肯普证明中的漏洞。
参考文献:
〔1〕、Holroyd,F.C.and Miller,R.G..The example that heawood shold have given Quart J Math.(1992). 43 (2),67-71
附英文版
Using H·Z-CP Solves Heawood Configuration
Zhang Yu-dian
Yu Xian Party School, Yu Xian 045100, Shanxi, China
Abstract: In this text, One Heawood configuration’s inevitable sets is found by using Heawoods-clouring procedure (abbreviated as H-CP) and Zhang Yu-dian clouring procedure (abbreviated as Z-CP), based on quantity and poison combination theory of coloring chain. And, one new procedure is found, which is named as H·Z-CP.
Key words: H-CP Z-CP H·Z-CP
Introduce
Thesis [1] made two main contributions to solving Heawood configuration. One is H-CP, by using it Heawood-coloring aperiodic transform’s Heawood configuration sets was found. The other one, in example II[1], provided Heawood-coloring periodic transform’s Heawood configuration. With it, Z-CP was found, and solved correct coloring for this configuration.
For the convenience of discuss, the simplest Heawood configuration model is given in [1] as follows.
As shown in Fig. 1, A, B,C ,D denote four colors, one roundlet denotes section V to be dyed, A1, B1, B2,C1 ,D1, denote five adjacent points border upon V, the pentagon area that forms is defined as pairs of B & A embedded area. Outside of V is A1-C1 chain and A1-D1 chain (because the head and trail is looped by V separately, so called loop, in order to distinguish with others). And there are B1-D2 chain and B 2-C2 chain also. A1, A2 is separated by C2-D2 chain. The other Heawood configuration is similar.
In this model, if add another coloring chain, many distinct normal triangle section map is formed(is G′). When to find the solution of map, it is found that distinct quantity combination and intersectant combination have effect on solution’s difference.
As follows, the detailed Heawood configuration’s inevitable sets is given.
Result
It is defined in latter figure as: a small transverse thread denotes a loop, a thick thread denotes a chain in which two or more coloring changed. B(D) etc. denotes that one point’s coloring is changed.
As shown in Fig. 2, if there are B1-A2 chain and D1-C2 chain in Fig. 1(can also be B2-A2 chain):
Its solution is: in A1-C1 loop, B and D is interchanged, a new A-D loop is formed (if it can’t be formed, belongs to another configuration). Then, C and B outside A-D loop is interchanged, and then V can be dyed with C color.
As shown in Fig. 3, if there are C1-D2 chain and D1-C2 chain in Fig. 1:
Its solution is: in A1-C1 loop, B and D is interchanged, a new B-C loop is formed, D and A outside B-C loop is interchanged, a new A-C loop is formed (if it can’t be formed, belongs to another configuration). Then, in A-C loop, B and D is interchanged, and then V can be dyed with B color.
As shown in Fig.4, if there are C1-D2 chain and B2-A2 chain in Fig. 1:
Its solution is: in A1-C1 loop, B and D is interchanged, a new B-C loop is formed, D and A outside B-C loop is interchanged, a new B-D loop is formed , in B-D loop, A and C is interchanged, a new B-C loop is formed, (if it can't be formed, belongs to another configuration). Then, in B-C loop, D and A is interchanged, and then V can be dyed with D color.
As shown in Fig.5, if B1-D2 chain and A1-D1 loop is intersectant in Fig. 4, new B1-A 3 loop and C1-A 3 loop are formed.
Its solution is:in A1-C1 loop, B and D is interchanged, a new B-C loop is formed, D and A outside B-C loop is interchanged, a new B-D loop is formed, in B-D loop, A and C is interchanged, a new A-D loop is formed, C and B outside A-D loop is interchanged, a new B-D loop is formed, (if it can't be formed, belongs to another configuration). Then, A and C outside B-D loop is interchanged, and then V can be dyed with A color.
As shown in Fig.6, if C1-D2 chain and A1-C1 loop is intersectant in Fig. 5, for simplicity, D can be dyed with B color in C1-D2 chain outside A1-C1 loop. See ○B in Fig.6.
Its solution is: in A1-C1 loop, B and D is interchanged, a new B-C loop is formed, D and A outside B-C loop is interchanged, a new B-D loop is formed, in B-D loop, A and C is interchanged, a new A-D loop is formed, C and B outside A-D loop is interchanged, a new A-C loop is formed, B and D outside A-C loop is interchanged, a new A-D loop is formed, (if it can't be formed, belongs to another configuration). Then, in A-D loop, C and B is interchanged, and then V can be dyed with C color.
As shown in Fig.7, if B1-D2 chain and B1-A3 loop is intersectant in Fig. 6, for simplicity, A can be dyed with C color in B1-A3 chain inside B1-D2 chain. See ○C in Fig. 7.
Its solution is: in A1-C1 loop, B and D is interchanged, a new B-C loop is formed, D and A outside B-C loop is interchanged, a new B-D loop is formed, in B-D loop, A and C is interchanged, a new A-D loop is formed, C and B outside A-D loop is interchanged, a new A-C loop is formed, B and D outside A-C loop is interchanged, a new B-C loop is formed, in B-C loop, D and A is interchanged, a new A-C loop is formed, (if it can't be formed, belongs to another configuration). Then, in A-C loop, B and D is interchanged, and then V can be dyed with B color.
As shown in Fig.8, if B1-D2 chain and C1-D2 chain is intersectant inside A1-C1 loop in Fig. 7.
Its solution is: in A1-C1 loop, B and D is interchanged, a new B-C loop is formed, D and A outside B-C loop is interchanged, a new B-D loop is formed, in B-D loop, A and C is interchanged, a new A-D loop is formed, C and B outside A-D loop is interchanged, a new A-C loop is formed, B and D outside A-C loop is interchanged, a new B-C loop is formed, in B-C loop, D and A is interchanged, a new B-D loop is formed, A and C outside B-D loop is interchanged, a new B-C loop is formed, (if it can't be formed, belongs to another configuration). Then, in B-C loop, D and A is interchanged, and then V can be dyed with D color.
As shown in Fig.8, if B2-A2 chain and A1-D2 loop is intersectant in Fig. 8.
Its solution is: in A1-C1 loop, B and D is interchanged, a new B-C loop is formed, D and A outside B-C loop is interchanged, a new B-D loop is formed, in B-D loop, A and C is interchanged, a new A-D loop is formed, C and B outside A-D loop is interchanged, a new A-C loop is formed, B and D outside A-C loop is interchanged, a new B-C loop is formed, in B-C loop, D and A is interchanged, a new B-D loop is formed, A and C outside B-D loop is interchanged, a new A-D loop is formed, in A-D loop, C and B is interchanged, a new B-D loop is formed, (if it can't be formed, belongs to another configuration). Then, in B-D loop, A and C is interchanged, and then V can be dyed with A color.
In Fig. 10, it is a ten-fold symmetrical Heawood configuration. Namely in Fig. 3, according intersectant combination method in Fig. 6,if C1-D2 chain and A1-C1 loop intersects, D1-C2 chain and A1-D1 loop intersects, D color point at C1-D2 chain outside A1-C1 loop and C color point at D1-C2 chain outside A1-D1 loop are both exchanged with B coloring, see ○B in Fig. 10. And then presume the exchanged C-B chain and D-B chain are symmetrically intersectant. This Heawood configuration is the topology transform form in example II [1].
For Fig. 10, if using the solution way in Fig. 9, 4 periodic transform’s Heawood configurations will come into being, and will be no result. But there is a common coloring character for the 4 sequence transform Heawood configurations, namely, they all contain A-B loop. And then, as follows Z-CP comes into being.
If Fig. 10(1) or 10(3) is known, firstly, C and D outside A-B loop interchanged, the new A-C loop and A-D loop(or B-C loop and B-D loop) come into being.then B(D) & B(C) (or A(D) & A(C)) interchange. The coloring number at the point of the pentagon is reducing to 3. Its conclusion is shown in Fig. 10(1) and Fig. 10(3).
If Fig. 10(2) or 10(4) is known, firstly, C and D outside A-B loop is interchanged, the new B-C (or A-D) chain come into being, then A(D) (or A(C)) at the side of B-C (or A-D) is interchange. The coloring number at the point of the pentagon is reducing to 3. Its conclusion is shown in Fig. 10(2) and Fig. 10(4).
The self-contained inevitable sets composed of Fig 2 to 10 will be proved as follows.
In the 4 color dyed G’, the quantity of distinct coloring chain formed by two colors in A, B,C ,D four colors have C42(=6) kinds totally. It is reflected in Heawood configuration, there are intersectant A1-C1 loop and A1-D1 loop whose start-point and end-point are all in center area. And there are B1-D2, B1-A2(B2-A2), B2-C2, C1-D2(D1-C2) 4 chains , whose start-point is in center area, and end-point is on the verge of the intersection area of A1-C1 loop and A1-D1 loop. There are 4 groups in total for the 4 kinds of chain’s distinct quantity combination in Heawood configuration:
B 1-A2、B 1-A2、B2-C2、B2-A2
B 1-A2、B 1-D2、B2-C2、D1-C2
C 1-D2、B 1-D2、B2-C2、B2-A2
C 1-D2、B 1-D2、B2-C2、D1-C2
There are C62(=15) kinds of two different situation’s combination in 6 kinds of chains, among them ,there are 3 kinds of not intersectant combinations:
A-B and C-D、A-C and B-D、A-D and B-C;
Otherwise there are 12 kinds of intersectant combinations:
A-B and A-C、A-D、B-C、B-D;
A-C and A-D、B-C、C-D ;
A-D and B-D、C-D;
B-C and B-D、C-D;
B-D and C-D。
Above 6 kinds of chain’s different quantity combinations(4 groups) and different situation combinations (intersectant 12 groups ) are two major variables, 16 kinds of Heawood configurations in different combination can be found totally. Then, on the “simplest structure” and “same solution” restrictive condition, verifiyed one by one, detailed conclusion is: Fig. 2 to Fig. 4 indicate 4 kinds of different quantity combinations. Among them, Fig. 2 indicates the former 2 groups. Fig. 5 to Fig. 9 indicate intersectant combination increased in turn. Among them, Fig. 9 contains12 kinds of intersectant combinations. Fig. 10 indicates specific quantity combinations sand intersectant combinations.
By this time, correct coloring for Heawood configuration is solved. The procedure which solve the problem, we name it H·Z-CP. The conclusion renovate the leak of kengpu proof.
Bibliography:
〔1〕、Holroyd,F.C.and Miller,R.G..The example that heawood shold have given Quart J Math.(1992). 43 (2),67-71
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[1] 李赤枫, 王 俊, 李 克, 等. 自生Mg2Si颗粒增强Al基复合材料的组织细化[J]. 中国有色金属学报, 2004, 14(2): 233-237.
[2] 殷 声. 燃烧合成[M]. 北京: 冶金工业出版社, 2004: 25-44.
[3] 王文新.大象征收过路费. 2006.5.21, /film/84974.htm.
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