http://codevs.cn/problem/1166/
区间dp， 手写大整数
1 #include<bits/stdc++.h>
2 #define mt(a,b) memset(a,b,sizeof(a))
3 using namespace std;
4 typedef long long LL;
5 const int M=1e2+10;
6
7 class BigInteger{
8 static const int M=1e2+10;
9 char s[M];
10 int length;
11 public:
12 void init(LL x){
13 if(x==0){
14 s[0]=‘0‘;
15 length=1;
16 return ;
17 }
18 length=0;
19 while(x){
20 s[length]=x%10+‘0‘;
21 x/=10;
22 length++;
23 }
24 }
25 bool operator <(const BigInteger &b) const{
26 if(length<b.length) return true;
27 if(length>b.length) return false;
28 for(int i=length-1;i>=0;i--){
29 if(s[i]<b.s[i]) return true;
30 if(s[i]>b.s[i]) return false;
31 }
32 return false;
33 }
34 BigInteger operator +(const BigInteger &b) const{
35 int buffer1[M];
36 int buffer2[M];
37 for(int i=0;i<M;i++){
38 buffer1[i]=0;
39 buffer2[i]=0;
40 }
41 for(int i=0;i<length;i++){
42 buffer1[i]=s[i]-‘0‘;
43 }
44 for(int i=0;i<b.length;i++){
45 buffer2[i]=b.s[i]-‘0‘;
46 }
47 int carry=0;
48 for(int i=0;i<M;i++){
49 int value=carry+buffer1[i]+buffer2[i];
50 buffer1[i]=value%10;
51 carry=value/10;
52 }
53 BigInteger result;
54 result.length=1;
55 for(int i=0;i<M;i++){
56 result.s[i]=buffer1[i]+‘0‘;
57 if(buffer1[i]) result.length=i+1;
58 }
59 return result;
60 }
61 BigInteger operator *(const BigInteger &b) const{
62 int buffer1[M];
63 int buffer2[M];
64 int buffer3[M];
65 for(int i=0;i<M;i++){
66 buffer1[i]=0;
67 buffer2[i]=0;
68 buffer3[i]=0;
69 }
70 for(int i=0;i<length;i++){
71 buffer1[i]=s[i]-‘0‘;
72 }
73 for(int i=0;i<b.length;i++){
74 buffer2[i]=b.s[i]-‘0‘;
75 }
76 for(int i=0;i<length;i++){
77 for(int j=0;j<b.length;j++){
78 buffer3[i+j]+=buffer1[i]*buffer2[j];
79 }
80 }
81 int carry=0;
82 for(int i=0;i<M;i++){
83 int value=carry+buffer3[i];
84 buffer3[i]=value%10;
85 carry=value/10;
86 }
87 BigInteger result;
88 result.length=1;
89 for(int i=0;i<M;i++){
90 result.s[i]=buffer3[i]+‘0‘;
91 if(buffer3[i]) result.length=i+1;
92 }
93 return result;
94 }
95 void output(){
96 for(int i=length-1;i>=0;i--){
97 putchar(s[i]);
98 }
99 }
100 };
101 int n,m;
102 LL a[M][M];
103 BigInteger dp[M][M];
104 BigInteger two[M];
105 bool flag[M][M];
106 void init(){
107 two[0].init(1);
108 BigInteger number;
109 number.init(2);
110 for(int i=1;i<M;i++){
111 two[i]=two[i-1]*number;
112 }
113 }
114 BigInteger dfs(int r,int L,int R){
115 BigInteger &d=dp[L][R];
116 if(flag[L][R]) return d;
117 flag[L][R]=true;
118 if(L==R){
119 BigInteger number;
120 number.init(a[r][L]);
121 d=number*two[m];
122 return d;
123 }
124 int time=m-(R-L+1)+1;
125 BigInteger number;
126 number.init(a[r][L]);
127 d=number*two[time]+dfs(r,L+1,R);
128 number.init(a[r][R]);
129 BigInteger buffer=number*two[time]+dfs(r,L,R-1);
130 if(d<buffer) d=buffer;
131 return d;
132 }
133 void solve(){
134 BigInteger answer;
135 answer.init(0);
136 for(int i=0;i<n;i++){
137 mt(flag,false);
138 answer=answer+dfs(i,0,m-1);
139 }
140 answer.output();
141 puts("");
142 }
143 int main(){
144 init();
145 while(~scanf("%d%d",&n,&m)){
146 for(int i=0;i<n;i++){
147 for(int j=0;j<m;j++){
148 scanf("%lld",&a[i][j]);
149 }
150 }
151 solve();
152 }
153 return 0;
154 }
http://codevs.cn/problem/3116/
大整数 计算
1 #include<bits/stdc++.h>
2 #define mt(a,b) memset(a,b,sizeof(a))
3 using namespace std;
4 typedef long long LL;
5 const int M=1e3+10;
6 class BigInteger{
7 static const int M=1e3+10;
8 char s[M];
9 int length;
10 public:
11 void init(char c[]){
12 int lc=strlen(c);
13 for(int i=0;i<lc;i++){
14 s[i]=c[lc-i-1];
15 }
16 length=lc;
17 }
18 void init(LL x){
19 if(x==0){
20 s[0]=‘0‘;
21 length=1;
22 return ;
23 }
24 length=0;
25 while(x){
26 s[length]=x%10+‘0‘;
27 x/=10;
28 length++;
29 }
30 }
31 BigInteger operator +(const BigInteger &b) const{
32 int buffer1[M];
33 int buffer2[M];
34 for(int i=0;i<M;i++){
35 buffer1[i]=0;
36 buffer2[i]=0;
37 }
38 for(int i=0;i<length;i++){
39 buffer1[i]=s[i]-‘0‘;
40 }
41 for(int i=0;i<b.length;i++){
42 buffer2[i]=b.s[i]-‘0‘;
43 }
44 int carry=0;
45 for(int i=0;i<M;i++){
46 int value=carry+buffer1[i]+buffer2[i];
47 buffer1[i]=value%10;
48 carry=value/10;
49 }
50 BigInteger result;
51 result.length=1;
52 for(int i=0;i<M;i++){
53 result.s[i]=buffer1[i]+‘0‘;
54 if(buffer1[i]) result.length=i+1;
55 }
56 return result;
57 }
58 BigInteger operator *(const BigInteger &b) const{
59 int buffer1[M];
60 int buffer2[M];
61 int buffer3[M];
62 for(int i=0;i<M;i++){
63 buffer1[i]=0;
64 buffer2[i]=0;
65 buffer3[i]=0;
66 }
67 for(int i=0;i<length;i++){
68 buffer1[i]=s[i]-‘0‘;
69 }
70 for(int i=0;i<b.length;i++){
71 buffer2[i]=b.s[i]-‘0‘;
72 }
73 for(int i=0;i<length;i++){
74 for(int j=0;j<b.length;j++){
75 buffer3[i+j]+=buffer1[i]*buffer2[j];
76 }
77 }
78 int carry=0;
79 for(int i=0;i<M;i++){
80 int value=carry+buffer3[i];
81 buffer3[i]=value%10;
82 carry=value/10;
83 }
84 BigInteger result;
85 result.length=1;
86 for(int i=0;i<M;i++){
87 result.s[i]=buffer3[i]+‘0‘;
88 if(buffer3[i]) result.length=i+1;
89 }
90 return result;
91 }
92 void output(){
93 for(int i=length-1;i>=0;i--){
94 putchar(s[i]);
95 }
96 }
97 }A,B,C;
98 char a[M];
99 char b[M];
100 int main(){
101 while(~scanf("%s%s",a,b)){
102 A.init(a);
103 B.init(b);
104 C=A+B;
105 C.output();
106 puts("");
107 }
108 return 0;
109 }
http://codevs.cn/problem/3117/
乘
1 #include<bits/stdc++.h>
2 #define mt(a,b) memset(a,b,sizeof(a))
3 using namespace std;
4 typedef long long LL;
5 const int M=1e3+10;
6 class BigInteger{
7 static const int M=1e3+10;
8 char s[M];
9 int length;
10 public:
11 void init(char c[]){
12 int lc=strlen(c);
13 for(int i=0;i<lc;i++){
14 s[i]=c[lc-i-1];
15 }
16 length=lc;
17 }
18 void init(LL x){
19 if(x==0){
20 s[0]=‘0‘;
21 length=1;
22 return ;
23 }
24 length=0;
25 while(x){
26 s[length]=x%10+‘0‘;
27 x/=10;
28 length++;
29 }
30 }
31 BigInteger operator +(const BigInteger &b) const{
32 int buffer1[M];
33 int buffer2[M];
34 for(int i=0;i<M;i++){
35 buffer1[i]=0;
36 buffer2[i]=0;
37 }
38 for(int i=0;i<length;i++){
39 buffer1[i]=s[i]-‘0‘;
40 }
41 for(int i=0;i<b.length;i++){
42 buffer2[i]=b.s[i]-‘0‘;
43 }
44 int carry=0;
45 for(int i=0;i<M;i++){
46 int value=carry+buffer1[i]+buffer2[i];
47 buffer1[i]=value%10;
48 carry=value/10;
49 }
50 BigInteger result;
51 result.length=1;
52 for(int i=0;i<M;i++){
53 result.s[i]=buffer1[i]+‘0‘;
54 if(buffer1[i]) result.length=i+1;
55 }
56 return result;
57 }
58 BigInteger operator *(const BigInteger &b) const{
59 int buffer1[M];
60 int buffer2[M];
61 int buffer3[M];
62 for(int i=0;i<M;i++){
63 buffer1[i]=0;
64 buffer2[i]=0;
65 buffer3[i]=0;
66 }
67 for(int i=0;i<length;i++){
68 buffer1[i]=s[i]-‘0‘;
69 }
70 for(int i=0;i<b.length;i++){
71 buffer2[i]=b.s[i]-‘0‘;
72 }
73 for(int i=0;i<length;i++){
74 for(int j=0;j<b.length;j++){
75 buffer3[i+j]+=buffer1[i]*buffer2[j];
76 }
77 }
78 int carry=0;
79 for(int i=0;i<M;i++){
80 int value=carry+buffer3[i];
81 buffer3[i]=value%10;
82 carry=value/10;
83 }
84 BigInteger result;
85 result.length=1;
86 for(int i=0;i<M;i++){
87 result.s[i]=buffer3[i]+‘0‘;
88 if(buffer3[i]) result.length=i+1;
89 }
90 return result;
91 }
92 void output(){
93 for(int i=length-1;i>=0;i--){
94 putchar(s[i]);
95 }
96 }
97 }A,B,C;
98 char a[M];
99 char b[M];
100 int main(){
101 while(~scanf("%s%s",a,b)){
102 A.init(a);
103 B.init(b);
104 C=A*B;
105 C.output();
106 puts("");
107 }
108 return 0;
109 }
http://codevs.cn/problem/3115/
减法 就 没 考虑进class了 ，那就复杂了 。
1 //#define txtout
2 //#define debug
3 #include<bits/stdc++.h>
4 #define mt(a,b) memset(a,b,sizeof(a))
5 using namespace std;
6 typedef long long LL;
7 const double pi=acos(-1.0);
8 const double eps=1e-8;
9 const int inf=0x3f3f3f3f;
10 const int M=1e3+10;
11 char a[M];
12 char b[M];
13 char answer[M];
14 void Reverse(char s[]){
15 int ls=strlen(s);
16 for(int i=0,j=ls-1;i<j;i++,j--){
17 swap(s[i],s[j]);
18 }
19 }
20 bool isBig(char s1[],char s2[]){
21 int l1=strlen(s1);
22 int l2=strlen(s2);
23 if(l1>l2) return true;
24 if(l1<l2) return false;
25 return strcmp(s1,s2)>0;
26 }
27 void solve(){
28 if(!strcmp(a,b)){
29 answer[0]=‘0‘;
30 answer[1]=0;
31 return ;
32 }
33 bool big=isBig(a,b);
34 if(!big) swap(a,b);
35 int la=strlen(a);
36 int lb=strlen(b);
37 Reverse(a);
38 Reverse(b);
39 for(int i=lb;i<la;i++){
40 b[i]=‘0‘;
41 }
42 int carry=0;
43 for(int i=0;i<la;i++){
44 if(a[i]-carry>=b[i]){
45 answer[i]=a[i]-carry-b[i]+‘0‘;
46 carry=0;
47 continue;
48 }
49 int need=1;
50 while(need*10+a[i]-carry<b[i]){
51 need++;
52 }
53 answer[i]=need*10+a[i]-carry-b[i]+‘0‘;
54 carry=need;
55 }
56 int len=0;
57 for(int i=la-1;i>=0;i--){
58 if(answer[i]>=‘1‘){
59 len=i+1;
60 break;
61 }
62 }
63 if(!big){
64 answer[len]=‘-‘;
65 len++;
66 }
67 answer[len]=0;
68 Reverse(answer);
69 }
70 int main(){
71 #ifdef txtout
72 #endif // txtout
73 while(~scanf("%s%s",a,b)){
74 solve();
75 puts(answer);
76 }
77 return 0;
78 }
end
时间： 2024-11-13 16:58:56