1 条题解
-
3
思路
- 先依据美味度排序
- 如果前两项口味不同,直接输出前两项的美味度之和
- 如果前两项的口味相同,先去比较其它不同口味的,如果按顺序第二个口味中,最美味的那个(称为A),与所有冰激凌最甜的那个(称为B),之和比第二美味的冰激淋(称为C)与A的和还大,输出A和B的甜度之和,否则输出B和C的甜度之和
加密代码
偏移量:
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#define ZAYq1URqHRwS7qAv082 a #define rUcWATN5o9L_RbLohv6 b #define j9YcgzjhMHaBQ begin #define UCOGjSPMa9WS_sVVEDB bool #define Y5Ozh0jGtcdjKk break #define bM9jCUglv4bH8GX cin #define qU4vZqjZI4E cmp #define NVntjB6iO30iuiWJE cout #define pm8VxjRHuL4nks df #define xQ10rk9BpoSkAWpD else #define tHVTXLW0WnKaFCb6iG1v end #define bRej_exfXKgHXjP false #define o4kVCNMscyHYUft1gI for #define WM60zMzeMZS1eL i #define Wfg2aCwTPU icecream #define W6Ekoo4zwG if #define SHDxMJMCO int #define iB8tp3911KbVJl8Bz ios #define lAt2xko4WC0GNfFl ll #define xhEjNLW2KC_i long #define Rnb2PVqqctjILPZ8N main #define y25ZaTyH max #define hq55zRaRAbMn5O3a4 n #define nsweGl7H namespace #define T_v3Yg5a5Fq3dPl return #define ML_ZL445 sort #define kvk7p4Ds std #define ZRfkOo9eLFK5 struct #define wQEv0cPuCshO_Mlb style #define IpaA76tS9ELtPl sweet #define K3p1Nez1 sync_with_stdio #define VEI0KIjYGR71T tie #define XSAfvxPGR using #define kpxPwunrXdvm8 v #define UzZfIwOFfd4fy vector #include<bits/stdc++.h> #define ll long long XSAfvxPGR nsweGl7H kvk7p4Ds; ZRfkOo9eLFK5 Wfg2aCwTPU { lAt2xko4WC0GNfFl wQEv0cPuCshO_Mlb,IpaA76tS9ELtPl; }; UCOGjSPMa9WS_sVVEDB qU4vZqjZI4E(Wfg2aCwTPU ZAYq1URqHRwS7qAv082,Wfg2aCwTPU rUcWATN5o9L_RbLohv6) { T_v3Yg5a5Fq3dPl ZAYq1URqHRwS7qAv082.IpaA76tS9ELtPl>rUcWATN5o9L_RbLohv6.IpaA76tS9ELtPl; } SHDxMJMCO Rnb2PVqqctjILPZ8N() { iB8tp3911KbVJl8Bz::K3p1Nez1(bRej_exfXKgHXjP); bM9jCUglv4bH8GX.VEI0KIjYGR71T(0); NVntjB6iO30iuiWJE.VEI0KIjYGR71T(0); lAt2xko4WC0GNfFl hq55zRaRAbMn5O3a4; bM9jCUglv4bH8GX>>hq55zRaRAbMn5O3a4; UzZfIwOFfd4fy<Wfg2aCwTPU> kpxPwunrXdvm8(hq55zRaRAbMn5O3a4+5); o4kVCNMscyHYUft1gI(lAt2xko4WC0GNfFl WM60zMzeMZS1eL=0;WM60zMzeMZS1eL<hq55zRaRAbMn5O3a4;WM60zMzeMZS1eL++) bM9jCUglv4bH8GX>>kpxPwunrXdvm8[WM60zMzeMZS1eL].wQEv0cPuCshO_Mlb>>kpxPwunrXdvm8[WM60zMzeMZS1eL].IpaA76tS9ELtPl; ML_ZL445(kpxPwunrXdvm8.j9YcgzjhMHaBQ(),kpxPwunrXdvm8.tHVTXLW0WnKaFCb6iG1v()-5,qU4vZqjZI4E); W6Ekoo4zwG(kpxPwunrXdvm8[0].wQEv0cPuCshO_Mlb!=kpxPwunrXdvm8[1].wQEv0cPuCshO_Mlb) NVntjB6iO30iuiWJE<<kpxPwunrXdvm8[0].IpaA76tS9ELtPl+kpxPwunrXdvm8[1].IpaA76tS9ELtPl; xQ10rk9BpoSkAWpD { lAt2xko4WC0GNfFl pm8VxjRHuL4nks=-1; o4kVCNMscyHYUft1gI(lAt2xko4WC0GNfFl WM60zMzeMZS1eL=2;WM60zMzeMZS1eL<hq55zRaRAbMn5O3a4;WM60zMzeMZS1eL++) { W6Ekoo4zwG(kpxPwunrXdvm8[0].wQEv0cPuCshO_Mlb!=kpxPwunrXdvm8[WM60zMzeMZS1eL].wQEv0cPuCshO_Mlb) { pm8VxjRHuL4nks=kpxPwunrXdvm8[0].IpaA76tS9ELtPl+kpxPwunrXdvm8[WM60zMzeMZS1eL].IpaA76tS9ELtPl; Y5Ozh0jGtcdjKk; } } NVntjB6iO30iuiWJE<<y25ZaTyH(kpxPwunrXdvm8[0].IpaA76tS9ELtPl+kpxPwunrXdvm8[1].IpaA76tS9ELtPl/2,pm8VxjRHuL4nks); } T_v3Yg5a5Fq3dPl 139; }
- 1
信息
- ID
- 102
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 3
- 标签
- 递交数
- 63
- 已通过
- 8
- 上传者