题目链接：http://codeforces.com/contest/552
题意：给你n个矩形，问你他们的面积和是多少，是这样的吧-w-
解：~
1 /* 2 * Problem: 3 * Author: SHJWUDP 4 * Created Time: 2015/7/3 星期五 12:19:42 5 * File Name: 233.cpp 6 * State: 7 * Memo: 8 */ 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 16 int n; 17 int main() { 18 #ifndef ONLINE_JUDGE 19 freopen("in", "r", stdin); 20 //freopen("out", "w", stdout); 21 #endif 22 while(~scanf("%d", &n)) { 23 int sum=0; 24 for(int i=0; i<n; i++) { 25 int a, b, c, d; 26 scanf("%d%d%d%d", &a, &b, &c, &d); 27 sum+=(c-a+1)*(d-b+1); 28 } 29 printf("%d\n", sum); 30 } 31 return 0; 32 }
题意：给你一个n，问你1~n有多少位数，tips：1,7,5是一位数；21,33,99是两位数
解：~
1 /* 2 * Problem: 3 * Author: SHJWUDP 4 * Created Time: 2015/7/3 星期五 12:19:42 5 * File Name: 233.cpp 6 * State: 7 * Memo: 8 */ 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 16 typedef long long int64; 17 18 const int MaxA=9+7; 19 20 int n; 21 int64 table[MaxA]; 22 int64 pow(int x, int n) { 23 int64 res=1; 24 while(n--) { 25 res*=x; 26 } 27 return res; 28 } 29 int main() { 30 #ifndef ONLINE_JUDGE 31 freopen("in", "r", stdin); 32 //freopen("out", "w", stdout); 33 #endif 34 table[0]=0; 35 for(int i=1; i<=9; i++) { 36 table[i]=table[i-1]+(pow(10, i)-pow(10, i-1))*i; 37 } 38 while(~scanf("%d", &n)) { 39 int x=0; 40 while(pow(10, x+1)<=n) x++; 41 printf("%I64d\n", table[x]+(x+1)*(n-pow(10, x)+1)); 42 } 43 return 0; 44 }
题意：给你一个天平和一些砝码，砝码的质量为w的0~100次方（每种一个），问你能不能称出质量为m的物品
解：看到一个很棒的解法，将m化为w进制，从低位到高位考虑，如果该位为0或者1，直接跳过，如果为w-1，将m增加1个当前位的砝码，继续运算，这样就绕过了情况较复杂的减法
1 /* 2 * Problem: 3 * Author: SHJWUDP 4 * Created Time: 2015/7/13 星期一 14:10:32 5 * File Name: 233.cpp 6 * State: 7 * Memo: 8 */ 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 16 const int MaxA=64+7; 17 18 int W, M; 19 int stk[MaxA], top; 20 int main() { 21 #ifndef ONLINE_JUDGE 22 freopen("in", "r", stdin); 23 //freopen("out", "w", stdout); 24 #endif 25 while(~scanf("%d%d", &W, &M)) { 26 if(W<=3) { 27 puts("YES"); 28 continue; 29 } 30 top=0; 31 bool ok=1; 32 do { 33 int x=M%W; 34 if(x<=1) M/=W; 35 else if(x==W-1) M=M/W+1; 36 else { 37 ok=0; 38 break; 39 } 40 } while(M); 41 puts(ok?"YES":"NO"); 42 } 43 return 0; 44 }
题意：给你欧式平面上的一些点，问你这些点做顶点能组成多少个三角形
解：~
1 /* 2 * Problem: 3 * Author: SHJWUDP 4 * Created Time: 2015/7/13 星期一 14:41:52 5 * File Name: 233.cpp 6 * State: 7 * Memo: 8 */ 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 #include <map> 14 15 using namespace std; 16 17 const int MaxA=2000+7; 18 19 struct Point { 20 int x, y; 21 } p[MaxA]; 22 23 int n; 24 int main() { 25 #ifndef ONLINE_JUDGE 26 freopen("in", "r", stdin); 27 //freopen("out", "w", stdout); 28 #endif 29 while(~scanf("%d", &n)) { 30 for(int i=0; i<n; i++) { 31 scanf("%d%d", &p[i].x, &p[i].y); 32 } 33 int ans=0; 34 for(int i=0; i<n; i++) { 35 int surplus=n-i-1; 36 map<pair<int, int>, int> M; 37 for(int j=i+1; j<n; j++) { 38 int dx=p[j].x-p[i].x; 39 int dy=p[j].y-p[i].y; 40 if(dx<0) dx=-dx, dy=-dy; 41 int d=__gcd(dx, dy); 42 dx/=d; dy/=d; 43 pair<int, int> x(dx, dy); 44 M[x]++; 45 } 46 for(const auto& u : M) { 47 surplus-=u.second; 48 ans+=u.second*surplus; 49 } 50 } 51 printf("%d\n", ans); 52 } 53 return 0; 54 }
题意：给你一个只有+和*还有数字1~9组成的算式，让你加入一对括号使得算式的值尽可能大
解：遇见这么深藏不露的模拟题也是醉了
1 /* 2 * Problem: 3 * Author: SHJWUDP 4 * Created Time: 2015/7/13 星期一 22:17:46 5 * File Name: 233.cpp 6 * State: 7 * Memo: 8 */ 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 16 typedef long long int64; 17 18 const int MaxA=5e3+7; 19 20 char str[MaxA]; 21 int64 arr[MaxA], n; 22 char op[MaxA]; //0:'+' 1:'*' 23 int64 func(int l, int r) { 24 int64 res=0, x=arr[l]; 25 for(int i=l; i<r; i++) { 26 if(op[i]=='+') { 27 res+=x; 28 x=arr[i+1]; 29 } else { 30 x*=arr[i+1]; 31 } 32 } 33 res+=x; 34 return res; 35 } 36 int main() { 37 #ifndef ONLINE_JUDGE 38 freopen("in", "r", stdin); 39 //freopen("out", "w", stdout); 40 #endif 41 while(~scanf("%s", str)) { 42 int len=strlen(str); 43 n=0; 44 vector<int> pos; 45 for(int i=0; i+1<len; i+=2) { 46 if(i==0 || str[i-1]!=str[i+1]) pos.push_back(n); 47 arr[n]=str[i]-'0'; 48 op[n++]=str[i+1]; 49 } 50 pos.push_back(n); 51 arr[n]=str[len-1]-'0'; 52 int64 ans=func(0, n); 53 for(int i=0, lim=(int)pos.size(); i<lim; i++) { 54 int64 x=0; 55 int64 pm=1; 56 int p=pos[i]-1; 57 while(p>=0 && op[p]=='*') pm*=arr[p--]; 58 if(0<=p) x=func(0, p); 59 for(int j=i+1; j<lim; j++) { 60 int64 y=0; 61 p=pos[j]; 62 int64 am=1; 63 while(p<n && op[p]=='*') am*=arr[++p]; 64 if(p+1<=n) y=func(p+1, n); 65 ans=max(ans, x+y+am*pm*func(pos[i], pos[j])); 66 } 67 } 68 printf("%I64d\n", ans); 69 } 70 return 0; 71 }
